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Infinite limits and asymptotes

Learning goals

  • Interprets and evaluates infinite limits and asymptotes using appropriate representations.
  • Uses graphs, tables, algebra, or known limit laws as appropriate.
  • Distinguishes finite, infinite, one-sided, and nonexistent behavior when relevant.
  • Explains conclusions using limit notation.

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Infinite limits and asymptotes 1
Let \(f(x)=\arctan x\). Find the limits of \(f(x)\) as \(x\to\infty\) and as \(x\to-\infty\). Then state all horizontal asymptotes of the graph.

Hints

- Think about the output values that the inverse tangent function can approach but never exceed. - Consider the two ends of the x-axis separately; they need not approach the same height.

Solution

1. The range of \(\arctan x\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\), and \(\arctan x\) approaches its upper endpoint as \(x\to\infty\). 2. Therefore \(\lim_{x\to\infty}\arctan x=\frac{\pi}{2}\), giving the horizontal asymptote \(y=\frac{\pi}{2}\) on the right. 3. As \(x\to-\infty\), \(\arctan x\to-\frac{\pi}{2}\), giving the horizontal asymptote \(y=-\frac{\pi}{2}\) on the left.

Answer

\(\lim_{x\to\infty}\arctan x=\frac{\pi}{2}\) and \(\lim_{x\to-\infty}\arctan x=-\frac{\pi}{2}\). The horizontal asymptotes are \(y=\frac{\pi}{2}\) and \(y=-\frac{\pi}{2}\).
Infinite limits and asymptotes 2
A function \(h\) satisfies \(\lim_{x\to-2^-}h(x)=+\infty\), \(\lim_{x\to-2^+}h(x)=+\infty\), \(\lim_{x\to\infty}h(x)=4\), and \(\lim_{x\to-\infty}h(x)=-1\). State every vertical and horizontal asymptote that is guaranteed by this information.

Hints

- Match unbounded behavior near a finite input with one type of asymptote. - Match finite limiting heights at the ends of the x-axis with another type of asymptote.

Solution

1. Both one-sided limits are unbounded as \(x\to-2\), so \(x=-2\) is a vertical asymptote. 2. Since \(h(x)\to4\) as \(x\to\infty\), \(y=4\) is a horizontal asymptote on the right. 3. Since \(h(x)\to-1\) as \(x\to-\infty\), \(y=-1\) is a horizontal asymptote on the left.

Answer

Vertical asymptote: \(x=-2\). Horizontal asymptotes: \(y=4\) as \(x\to\infty\) and \(y=-1\) as \(x\to-\infty\).
Infinite limits and asymptotes 3
Find the vertical asymptote of \(f\) from the graph. Write the two one-sided limits there and explain their signs.
Figure for problem 562458

Hints

- Locate the input where the branches climb sharply. - Compare the direction of the left and right branches. - A vertical asymptote is written as an equation using \(x\), not \(y\).

Solution

1. Both branches rise without bound as \(x\) approaches \(2\). 2. The vertical asymptote is \(x=2\). 3. Both \(\lim_{x\to2^-}f(x)\) and \(\lim_{x\to2^+}f(x)\) equal \(+\infty\) in the extended-limit notation.

Answer

Vertical asymptote \(x=2\); both one-sided limits are \(+\infty\).
Infinite limits and asymptotes 4
As \(x\) moves far to the right, what horizontal line does \(f\) appear to approach? Write its equation and explain whether \(f\) has to reach that line.
Figure for problem 562459

Hints

- Follow the rightmost part of the curve. - What height do the graph values settle near? - Would reaching a line at one input change what happens far to the right?

Solution

1. The plotted curve gets closer to height \(2\) toward the right edge, suggesting the same trend for larger inputs. 2. The graph therefore suggests the horizontal asymptote \(y=2\). 3. The curve stays above that height in the plotted range. A horizontal asymptote does not require a curve to reach or cross the line. A finite graph alone cannot prove its end behavior.

Answer

The graph suggests the horizontal asymptote \(y=2\). A curve need not touch its horizontal asymptote.
Infinite limits and asymptotes 5
Let \(f(x)=\frac{5}{x+2}\). As \(x\to-2^+\), does \(f(x)\) approach \(+\infty\) or \(-\infty\)? Explain using the signs of the numerator and denominator.

Hints

- Check whether \(x+2\) is positive or negative just right of \(-2\). - The numerator stays fixed and positive. - Think about division by very small positive numbers.

Solution

1. The numerator is positive. 2. Approaching \(-2\) from the right makes \(x+2\) positive and arbitrarily small. 3. Dividing \(5\) by smaller positive values makes the quotient grow without bound, so the right-hand infinite limit is \(+\infty\).

Answer

\(+\infty\); the denominator approaches zero through positive values.
Infinite limits and asymptotes 6
Match each limit statement to the type of asymptote it describes. I. \(\lim_{x\to\infty}g(x)=0\) II. \(\lim_{x\to1^+}g(x)=+\infty\) State the equation of the asymptote in each case.

Hints

- Decide whether the input or the function's output is tending to infinity. - A horizontal line has an equation using \(y\). - A vertical line has an equation using \(x\).

Solution

1. In I, the input grows without bound while the output approaches \(0\), giving a horizontal asymptote \(y=0\). 2. In II, the input approaches the finite value \(1\) while output grows without bound, giving a vertical asymptote \(x=1\).

Answer

I. Horizontal asymptote \(y=0\). II. Vertical asymptote \(x=1\).
Infinite limits and asymptotes 7
Evaluate each one-sided limit. a) \(\lim_{x\to1^-}\frac{x+2}{x-1}\) b) \(\lim_{x\to-2^+}\frac{3}{x+2}\)

Hints

- Determine the sign of the numerator and denominator near the approach value. - Track whether the denominator approaches zero from the positive or negative side.

Solution

1. In part a, the numerator approaches \(3\), while the denominator approaches \(0\) through negative values. Therefore, the quotient approaches \(-\infty\). 2. In part b, the numerator is positive and the denominator approaches \(0\) through positive values. Therefore, the quotient approaches \(+\infty\).

Answer

a) \(-\infty\) b) \(+\infty\)
Infinite limits and asymptotes 8
Find the limits as \(x\to\infty\) and as \(x\to-\infty\). a) \(f(x)=5-\frac{4}{x^2}\) b) \(g(x)=\frac{3(4^x)+2}{4^x}\) c) \(h(x)=8(0.5)^x-2\)

Hints

- Rewrite quotients when possible. - Recall the behavior of exponential functions with bases greater than \(1\) and between \(0\) and \(1\).

Solution

1. In part a, \(\frac{4}{x^2}\to0\) in both directions, so \(f(x)\to5\). 2. Rewrite part b as \(g(x)=3+\frac{2}{4^x}\). As \(x\to\infty\), the fraction approaches \(0\), so \(g(x)\to3\). As \(x\to-\infty\), \(4^x\to0^+\), so the fraction approaches \(+\infty\), and \(g(x)\to+\infty\). 3. In part c, \((0.5)^x\to0\) as \(x\to\infty\), so \(h(x)\to-2\). As \(x\to-\infty\), \((0.5)^x\to+\infty\), so \(h(x)\to+\infty\).

Answer

a) \(\lim_{x\to\pm\infty}f(x)=5\) b) \(\lim_{x\to\infty}g(x)=3\); \(\lim_{x\to-\infty}g(x)=+\infty\) c) \(\lim_{x\to\infty}h(x)=-2\); \(\lim_{x\to-\infty}h(x)=+\infty\)
Infinite limits and asymptotes 9
Let \(f(x)=\frac{8x+12}{4x-5}\). Find \(\lim_{x\to\infty}f(x)\), and briefly explain its geometric meaning for the graph.

Hints

- Divide by the highest power of \(x\). - Connect a finite limit at infinity with a horizontal asymptote.

Solution

1. Divide the numerator and denominator by \(x\): \(f(x)=\frac{8+\frac{12}{x}}{4-\frac{5}{x}}\). 2. As \(x\to\infty\), the reciprocal terms approach \(0\), so the limit is \(\frac{8}{4}=2\). 3. Geometrically, the graph approaches the horizontal line \(y=2\) for large positive \(x\).

Answer

The limit is \(2\), so \(y=2\) is a horizontal asymptote.
Infinite limits and asymptotes 10
Analyze \(f(x)=\frac{x^2-1}{x^2-2x+1}\) at \(x=1\). Evaluate \(\lim_{x\to 1}f(x)\), or explain why it does not exist, and determine whether \(x=1\) is a removable discontinuity or a vertical asymptote.

Hints

- Factor the numerator and denominator first. - Can the rational expression be simplified? - What happens when the numerator approaches a nonzero number while the denominator approaches zero? - Compare the sign of the denominator on the two sides of \(x=1\).

Solution

1. Factor the numerator and denominator: \(x^2-1=(x-1)(x+1)\) and \(x^2-2x+1=(x-1)^2\). 2. For \(x\neq 1\), simplify: \(f(x)=\frac{x+1}{x-1}\). 3. As \(x\to 1\), the numerator approaches \(2\), while the denominator approaches \(0\). 4. The left-hand limit is \(\lim_{x\to 1^-}f(x)=-\infty\), and the right-hand limit is \(\lim_{x\to 1^+}f(x)=+\infty\). 5. Because the one-sided limits are different, the two-sided limit does not exist. The discontinuity is infinite, and the vertical asymptote is \(x=1\).

Answer

\(\lim_{x\to 1}f(x)\) does not exist because \(\lim_{x\to 1^-}f(x)=-\infty\) and \(\lim_{x\to 1^+}f(x)=+\infty\). Therefore, \(x=1\) is a vertical asymptote, not a removable discontinuity.
Infinite limits and asymptotes 11
Let \(f(x)=(x^2+3x)\ln(x)\) for \(x>0\). You may use the known limit \(\lim_{x\to0^+}x\ln(x)=0\). Find \(\lim_{x\to0^+}f(x)\). Is \(x=0\) a vertical asymptote of its graph? Explain why \(\ln(x)\to-\infty\) does not mean the product also becomes unbounded.

Hints

- Can you rearrange the product so the supplied limit appears in it? - A product of a small factor and an unbounded factor needs more care than substitution. - An asymptote at a finite input requires unbounded nearby function values.

Solution

1. Rewrite the product as \(f(x)=(x+3)(x\ln(x))\). 2. As \(x\to0^+\), the first factor approaches \(3\) and the second approaches \(0\) by the supplied limit, so \(\lim_{x\to0^+}f(x)=0\). 3. The product has a finite limit despite the unbounded logarithm. Therefore \(x=0\) is not a vertical asymptote.

Answer

The limit is \(0\), so \(x=0\) is not a vertical asymptote. The factor approaching zero changes the behavior of the product.
Infinite limits and asymptotes 12
Consider the family of functions \(f_n(x)=\frac{1}{(x-5)^n}\), where \(n\in\mathbb{N}\setminus\{0\}\). 1. Suppose \(n\) is odd. Find \(\lim_{x\to 5^+}f_n(x)\) and \(\lim_{x\to 5^-}f_n(x)\). 2. Suppose \(n\) is even. Explain why \(f_n(x)>0\) for every input in its domain, and find \(\lim_{x\to 5}f_n(x)\).

Hints

- What sign results when a negative number is raised to an odd power? - Is \(x-5\) positive or negative just to the left of \(5\)? - What sign does any nonzero real number have after it is raised to an even power?

Solution

1. If \(n\) is odd and \(x\to 5^+\), then \(x-5\to 0^+\), so \((x-5)^n\to 0^+\). Therefore, \(\lim_{x\to 5^+}f_n(x)=+\infty\). 2. If \(n\) is odd and \(x\to 5^-\), then \(x-5\to 0^-\). An odd power remains negative, so \((x-5)^n\to 0^-\). Therefore, \(\lim_{x\to 5^-}f_n(x)=-\infty\). 3. If \(n\) is even, then \((x-5)^n>0\) for every \(x\neq 5\). Since the numerator is also positive, \(f_n(x)>0\) throughout its domain. 4. As \(x\to 5\), the even-powered denominator approaches \(0^+\) from both sides. Thus, \(\lim_{x\to 5}f_n(x)=+\infty\).

Answer

1. For odd \(n\), \(\lim_{x\to 5^+}f_n(x)=+\infty\) and \(\lim_{x\to 5^-}f_n(x)=-\infty\). 2. For even \(n\), \((x-5)^n>0\) whenever \(x\neq 5\), so \(f_n(x)>0\). Also, \(\lim_{x\to 5}f_n(x)=+\infty\).
Infinite limits and asymptotes 13
Consider \(g_k(x)=\frac{-2}{(x+1)^k}\), where \(k\in\mathbb{N}\setminus\{0\}\). Find \(\lim_{x\to -1^-}g_k(x)\) for odd values of \(k\) and for even values of \(k\). Explain how the negative numerator affects the sign of each limit.

Hints

- Determine the sign of the denominator when \(x<-1\), considering whether \(k\) is odd or even. - How does the numerator \(-2\) affect the sign of the quotient? - Separate the sign of the denominator from the sign of the entire fraction.

Solution

1. As \(x\to -1^-\), the expression \(x+1\) is negative and approaches \(0\). 2. If \(k\) is odd, then \((x+1)^k\to 0^-\). Dividing the negative numerator \(-2\) by a negative denominator gives a positive result, so \(\lim_{x\to -1^-}g_k(x)=+\infty\). 3. If \(k\) is even, then \((x+1)^k\to 0^+\). Dividing \(-2\) by a positive denominator gives a negative result, so \(\lim_{x\to -1^-}g_k(x)=-\infty\). 4. Compared with a positive numerator, the negative numerator reverses the sign of the function values.

Answer

For odd \(k\), \(\lim_{x\to -1^-}g_k(x)=+\infty\). For even \(k\), \(\lim_{x\to -1^-}g_k(x)=-\infty\). The negative numerator reverses the sign determined by the denominator.
Infinite limits and asymptotes 14
Consider \(g(x)=\frac{1}{\ln x}\). Identify its vertical asymptote and determine the one-sided behavior there. Then find its horizontal asymptote as \(x\to\infty\).

Hints

- First find where the denominator can approach zero within the logarithm's domain. - The sign of the logarithm changes as its input passes through \(1\). - For end behavior, ask what happens to the reciprocal when the denominator grows without bound.

Solution

1. A vertical asymptote can occur where \(\ln x=0\), which is at \(x=1\). 2. For \(0<x<1\), \(\ln x<0\) and approaches \(0\) from below as \(x\to1^-\), so \(g(x)\to-\infty\). 3. For \(x>1\), \(\ln x>0\) and approaches \(0\) from above as \(x\to1^+\), so \(g(x)\to+\infty\). Thus \(x=1\) is a vertical asymptote. 4. As \(x\to\infty\), \(\ln x\to\infty\), so \(\frac{1}{\ln x}\to0\). Therefore \(y=0\) is the horizontal asymptote on the right.

Answer

Vertical asymptote: \(x=1\), with \(\lim_{x\to1^-}g(x)=-\infty\) and \(\lim_{x\to1^+}g(x)=+\infty\). Horizontal asymptote as \(x\to\infty\): \(y=0\).
Infinite limits and asymptotes 15
A function is defined by \(p(x)=\frac{e^x}{1+e^x}\). Find its end behavior as \(x\to\infty\) and \(x\to-\infty\), and state the horizontal asymptote associated with each end.

Hints

- At the positive end, rewrite the fraction so the exponential no longer grows in both numerator and denominator. - At the negative end, use the limiting behavior of \(e^x\).

Solution

1. As \(x\to\infty\), divide numerator and denominator by \(e^x\): \(p(x)=\frac{1}{e^{-x}+1}\to1\). 2. Therefore the graph approaches \(y=1\) as \(x\to\infty\). 3. As \(x\to-\infty\), \(e^x\to0\), so \(p(x)=\frac{e^x}{1+e^x}\to0\). 4. Therefore the graph approaches \(y=0\) as \(x\to-\infty\).

Answer

\(\lim_{x\to\infty}p(x)=1\), so \(y=1\) is the right-end horizontal asymptote. \(\lim_{x\to-\infty}p(x)=0\), so \(y=0\) is the left-end horizontal asymptote.
Infinite limits and asymptotes 16
Consider \(h(x)=\frac{\ln x}{x}\), defined for \(x>0\). Analyze the graph near the left boundary of its domain and as \(x\to\infty\). State any vertical or horizontal asymptotes that follow from these limits.

Hints

- Treat the boundary \(x=0\) using only inputs from the function's domain. - Compare the long-run growth rates of a logarithm and a linear function.

Solution

1. As \(x\to0^+\), \(\ln x\to-\infty\) while \(x\) remains positive and approaches \(0\). Thus \(\frac{\ln x}{x}\to-\infty\). 2. Therefore \(x=0\) is a vertical asymptote, approached from the right. 3. As \(x\to\infty\), logarithmic growth is slower than linear growth, so \(\frac{\ln x}{x}\to0\). 4. Therefore \(y=0\) is a horizontal asymptote as \(x\to\infty\).

Answer

\(\lim_{x\to0^+}h(x)=-\infty\), so \(x=0\) is a vertical asymptote. \(\lim_{x\to\infty}h(x)=0\), so \(y=0\) is a horizontal asymptote on the right.
Infinite limits and asymptotes 17
Let \(q(x)=xe^{-x}\). Determine the end behavior of \(q\) as \(x\to\infty\) and as \(x\to-\infty\). State whether the graph has a horizontal asymptote on either end.

Hints

- Rewrite the positive-end expression as a quotient to compare growth rates. - At the negative end, track both the sign of \(x\) and the behavior of the exponential factor.

Solution

1. As \(x\to\infty\), \(q(x)=\frac{x}{e^x}\), and exponential growth dominates linear growth, so \(q(x)\to0\). 2. Therefore \(y=0\) is a horizontal asymptote as \(x\to\infty\). 3. As \(x\to-\infty\), the factor \(x\) is negative with unbounded magnitude and \(e^{-x}\to\infty\), so their product tends to \(-\infty\). 4. Since the left-end limit is not finite, there is no horizontal asymptote as \(x\to-\infty\).

Answer

\(\lim_{x\to\infty}q(x)=0\), so \(y=0\) is a right-end horizontal asymptote. \(\lim_{x\to-\infty}q(x)=-\infty\), so there is no left-end horizontal asymptote.
Infinite limits and asymptotes 18
Let \(f(x)=\frac{x^3+1}{x^2+1}\). Determine whether the graph has any vertical asymptotes. Then find a slant asymptote by rewriting the function so that the difference between \(f(x)\) and a line is visible.

Hints

- Check the denominator before assuming a rational function has a vertical asymptote. - Divide the polynomials and examine the leftover term at large positive and negative inputs.

Solution

1. Since \(x^2+1>0\) for every real \(x\), the denominator never vanishes, so there are no vertical asymptotes. 2. Polynomial division gives \(f(x)=x+\frac{1-x}{x^2+1}\). 3. As \(x\to\infty\) or \(x\to-\infty\), \(\frac{1-x}{x^2+1}\to0\). 4. Therefore \(f(x)-x\to0\) on both ends, so the slant asymptote is \(y=x\).

Answer

There are no vertical asymptotes. The slant asymptote is \(y=x\) as \(x\to\pm\infty\).
Infinite limits and asymptotes 19
Let \(f(x)=\frac{1}{\sqrt{5-x}}\), with its largest real domain. Determine the domain, describe the behavior as \(x\to5^-\), and find the end behavior as \(x\to-\infty\). State the asymptotes supported by these limits.

Hints

- The square root is in the denominator, so its radicand must be strictly positive. - A vertical asymptote can be established by unbounded behavior from just one side. - For the far-left behavior, track what happens to the denominator's size.

Solution

1. The denominator requires \(5-x>0\), so the domain is \((-\infty, 5)\). 2. As \(x\to5^-\), \(\sqrt{5-x}\to0^+\), so \(f(x)\to+\infty\). Thus \(x=5\) is a vertical asymptote even though the function exists only on one side of it. 3. As \(x\to-\infty\), \(5-x\to\infty\), so \(\sqrt{5-x}\to\infty\) and \(f(x)\to0\). 4. Therefore \(y=0\) is a horizontal asymptote on the left.

Answer

Domain: \((-\infty, 5)\). \(\lim_{x\to5^-}f(x)=+\infty\), so \(x=5\) is a vertical asymptote. \(\lim_{x\to-\infty}f(x)=0\), so \(y=0\) is a left-end horizontal asymptote.
Infinite limits and asymptotes 20
The table compares two ratios for increasingly large positive \(x\). <table><tr><th>\(x\)</th><td>\(10\)</td><td>\(20\)</td><td>\(40\)</td><td>\(80\)</td></tr><tr><th>\(\frac{x^3}{2^x}\)</th><td>\(0.9766\)</td><td>\(0.007629\)</td><td>\(5.82\times10^{-8}\)</td><td>\(4.24\times10^{-19}\)</td></tr><tr><th>\(\frac{\ln x}{x}\)</th><td>\(0.2303\)</td><td>\(0.1498\)</td><td>\(0.0922\)</td><td>\(0.0548\)</td></tr></table> a) What end-behavior trend does each ratio suggest? b) For each ratio, identify which part is growing faster: \(2^x\) or \(x^3\), and \(x\) or \(\ln x\). c) Based on these trends, what horizontal asymptotes would you expect for \(A(x)=5+\frac{x^3}{2^x}\) and \(B(x)=\frac{\ln x}{x}\) as \(x\to\infty\)? Explain the limitation of using a finite table as evidence.

Hints

- Interpret each displayed ratio as a comparison of the numerator's growth with the denominator's growth. - Ask what it means when a positive ratio becomes tiny as \(x\) grows. - For each transformed function, connect the ratio's apparent end behavior to the constant value approached by the whole expression.

Solution

1. The values of \(x^3/2^x\) rapidly decrease toward \(0\), while the values of \(\ln x/x\) also decrease toward \(0\). 2. A ratio tending toward \(0\) indicates that its denominator is outgrowing its numerator: the table therefore suggests that \(2^x\) outgrows \(x^3\), and \(x\) outgrows \(\ln x\). 3. If those end-behavior trends continue, \(A(x)\to5\), so the expected horizontal asymptote is \(y=5\). Also \(B(x)\to0\), so the expected horizontal asymptote is \(y=0\). 4. The finite table provides numerical evidence and motivation for these relative-growth facts; by itself it is not a proof of behavior for arbitrarily large \(x\).

Answer

a) Both ratios suggest a limit of \(0\) as \(x\to\infty\). b) The data suggest that \(2^x\) grows faster than \(x^3\), and \(x\) grows faster than \(\ln x\). c) Expected asymptotes: \(A\): \(y=5\); \(B\): \(y=0\). A finite table supports these conclusions numerically but does not prove the limiting behavior for all sufficiently large \(x\).
Infinite limits and asymptotes 21
A function has the form \(h(x)=a+\frac{3}{x-4}\), with \(x\ne4\). Choose \(a\) so its graph has horizontal asymptote \(y=-2\). What is its vertical asymptote? Explain both from the formula.

Hints

- What happens to the fraction when \(|x|\) becomes very large? - Which part of the expression sets the final height? - Look for the excluded input where the denominator vanishes.

Solution

1. As \(x\to\pm\infty\), the fraction \(3/(x-4)\) approaches \(0\), so the horizontal asymptote is \(y=a\). 2. Set \(a=-2\). 3. The denominator approaches zero at \(x=4\) while the numerator stays nonzero, giving the vertical asymptote \(x=4\).

Answer

\(a=-2\); vertical asymptote \(x=4\).
Infinite limits and asymptotes 22
A tethered balloon's height after \(t\) minutes is modeled by \(H(t)=80(1-e^{-t/5})\) feet for \(t\ge0\). Find its height as \(t\to\infty\), give the horizontal asymptote, and explain what that line means in this model.

Hints

- Consider what happens to \(e^{-t/5}\) for very large positive \(t\). - What number does the whole formula approach? - Interpret the limiting height in the story.

Solution

1. As \(t\to\infty\), the exponent \(-t/5\to-\infty\), so \(e^{-t/5}\to0\). 2. Thus \(H(t)\to80(1-0)=80\) feet. 3. The horizontal asymptote is \(H=80\); the modeled height approaches \(80\) feet rather than increasing forever.

Answer

Limit \(80\,\text{ft}\); horizontal asymptote \(H=80\). The balloon levels off near that height.
Infinite limits and asymptotes 23
Consider the rational function \(f(x)=\frac{x^2-3x}{x^2-9}\). a) State the domain of \(f\) and classify each discontinuity as removable or infinite. b) Give the equation of the vertical asymptote. c) Use one-sided limits to describe the behavior of \(f(x)\) near the vertical asymptote.

Hints

- First find the values that make the denominator zero. - Factor and simplify the rational expression before classifying the discontinuities. - What does a common numerator and denominator factor imply about the discontinuity? - Check the sign of the simplified expression on each side of the vertical asymptote.

Solution

1. Factor the denominator: \(x^2-9=(x-3)(x+3)\). Therefore, \(D_f=\mathbb{R}\setminus\{-3, 3\}\). 2. Factor the numerator and simplify for \(x\neq 3\): \(f(x)=\frac{x(x-3)}{(x-3)(x+3)}=\frac{x}{x+3}\). 3. The factor \(x-3\) cancels, and \(\lim_{x\to 3}\frac{x}{x+3}=\frac{1}{2}\). Thus, \(x=3\) is a removable discontinuity. 4. The factor \(x+3\) remains in the denominator, so \(x=-3\) is an infinite discontinuity and the vertical asymptote is \(x=-3\). 5. As \(x\to -3^-\), the numerator is negative and the denominator approaches \(0^-\), so \(\lim_{x\to -3^-}f(x)=+\infty\). 6. As \(x\to -3^+\), the numerator is negative and the denominator approaches \(0^+\), so \(\lim_{x\to -3^+}f(x)=-\infty\).

Answer

a) \(D_f=\mathbb{R}\setminus\{-3, 3\}\). The discontinuity at \(x=3\) is removable, and the discontinuity at \(x=-3\) is infinite. b) \(x=-3\) c) \(\lim_{x\to -3^-}f(x)=+\infty\) and \(\lim_{x\to -3^+}f(x)=-\infty\).
Infinite limits and asymptotes 24
Let \(h(x)=\frac{5-x}{x^2-10x+25}\). a) State the domain of \(h\). b) Show algebraically that the discontinuity is a vertical asymptote. c) Find the one-sided limits of \(h(x)\) as \(x\) approaches \(5\).

Hints

- Rewrite the denominator as a perfect-square binomial. - Pay close attention to the negative sign when rewriting the numerator. - If a denominator factor remains after complete simplification, the excluded value is a vertical asymptote. - Determine the sign of \(x-5\) just to the left and just to the right of \(5\).

Solution

1. Factor the denominator: \(x^2-10x+25=(x-5)^2\). Therefore, the domain is \(D_h=\mathbb{R}\setminus\{5\}\). 2. Rewrite the numerator as \(5-x=-(x-5)\). For \(x\neq 5\), \(h(x)=\frac{-(x-5)}{(x-5)^2}=-\frac{1}{x-5}\). A denominator factor remains after simplification, so \(x=5\) is a vertical asymptote. 3. As \(x\to 5^-\), \(x-5\to 0^-\), so \(-\frac{1}{x-5}\to +\infty\). Thus, \(\lim_{x\to 5^-}h(x)=+\infty\). 4. As \(x\to 5^+\), \(x-5\to 0^+\), so \(-\frac{1}{x-5}\to -\infty\). Thus, \(\lim_{x\to 5^+}h(x)=-\infty\).

Answer

a) \(D_h=\mathbb{R}\setminus\{5\}\) b) For \(x\neq 5\), \(h(x)=-\frac{1}{x-5}\). Since a denominator factor remains, \(x=5\) is a vertical asymptote. c) \(\lim_{x\to 5^-}h(x)=+\infty\) and \(\lim_{x\to 5^+}h(x)=-\infty\).
Infinite limits and asymptotes 25
Let \(g(x)=1-\frac{9}{x^2-16}\). a) State the domain of \(g\). b) Find \(\lim_{x\to -4^-}g(x)\), \(\lim_{x\to -4^+}g(x)\), \(\lim_{x\to 4^-}g(x)\), and \(\lim_{x\to 4^+}g(x)\). c) Use a limit as \(x\to\infty\) to determine the horizontal asymptote of the graph.

Hints

- For which values is the denominator zero? - Near each excluded value, determine whether the denominator approaches \(0^+\) or \(0^-\). - What happens to the fraction as \(x\) becomes very large?

Solution

1. The denominator is zero when \(x^2-16=0\), so \(x=\pm 4\). Therefore, \(D_g=\mathbb{R}\setminus\{-4, 4\}\). 2. As \(x\to 4^-\), \(x^2-16\to 0^-\). Thus, \(\frac{9}{x^2-16}\to -\infty\), and \(g(x)\to +\infty\). 3. As \(x\to 4^+\), \(x^2-16\to 0^+\). Thus, \(\frac{9}{x^2-16}\to +\infty\), and \(g(x)\to -\infty\). 4. Because \(g\) is even, the behavior at \(x=-4\) is reflected across the y-axis. Therefore, \(\lim_{x\to -4^-}g(x)=-\infty\) and \(\lim_{x\to -4^+}g(x)=+\infty\). 5. As \(x\to\infty\), \(\frac{9}{x^2-16}\to 0\), so \(\lim_{x\to\infty}g(x)=1\). The horizontal asymptote is \(y=1\).

Answer

a) \(D_g=\mathbb{R}\setminus\{-4, 4\}\) b) \(\lim_{x\to -4^-}g(x)=-\infty\), \(\lim_{x\to -4^+}g(x)=+\infty\), \(\lim_{x\to 4^-}g(x)=+\infty\), and \(\lim_{x\to 4^+}g(x)=-\infty\) c) \(y=1\)
Infinite limits and asymptotes 26
Determine the end behavior of \(f(x)=(x^3+1)e^{-x}\) as \(x\to\infty\) and as \(x\to-\infty\).

Hints

- Analyze the two factors separately. - Compare the growth rate of a polynomial with the growth or decay rate of an exponential function. - Track the sign of each factor for large negative values of \(x\).

Solution

1. As \(x\to\infty\), the exponential decay of \(e^{-x}\) is faster than the polynomial growth of \(x^3+1\). Therefore, \((x^3+1)e^{-x}\to0\). 2. As \(x\to-\infty\), \(x^3+1\to-\infty\) and \(e^{-x}\to\infty\). The product is negative with unbounded magnitude, so \(f(x)\to-\infty\).

Answer

\(\lim_{x\to\infty}f(x)=0\) and \(\lim_{x\to-\infty}f(x)=-\infty\)
Infinite limits and asymptotes 27
Determine the end behavior of \(g(x)=2x^2-e^{0.1x}\) as \(x\to\infty\) and as \(x\to-\infty\).

Hints

- Identify which term has the faster growth rate for large positive \(x\). - Factoring out the exponential term can make the dominant behavior clearer. - Determine what \(e^{0.1x}\) does as \(x\to-\infty\).

Solution

1. As \(x\to\infty\), the exponential term grows faster than the polynomial term. Write \(g(x)=e^{0.1x}\left(2x^2e^{-0.1x}-1\right)\). Since \(2x^2e^{-0.1x}\to0\), the expression in parentheses approaches \(-1\), so \(g(x)\to-\infty\). 2. As \(x\to-\infty\), \(2x^2\to\infty\) while \(e^{0.1x}\to0\). Therefore, \(g(x)\to\infty\).

Answer

\(\lim_{x\to\infty}g(x)=-\infty\) and \(\lim_{x\to-\infty}g(x)=\infty\)
Infinite limits and asymptotes 28
Let \(f(x)=\frac{100x^2+e^x}{e^x-x^4}\). Find \(\lim_{x\to\infty}f(x)\) and \(\lim_{x\to-\infty}f(x)\). Justify each result by comparing the growth rates of the terms.

Hints

- For large positive \(x\), divide by the fastest-growing term. - Recall that \(e^x\) grows faster than any power of \(x\). - For large negative \(x\), determine what happens to \(e^x\). - Analyze the two directions separately.

Solution

1. As \(x\to\infty\), divide the numerator and denominator by \(e^x\): \(f(x)=\frac{100x^2/e^x+1}{1-x^4/e^x}\). Since every polynomial grows more slowly than \(e^x\), both polynomial-to-exponential ratios approach \(0\). Therefore, \(\lim_{x\to\infty}f(x)=1\). 2. As \(x\to-\infty\), \(e^x\to0\), so the polynomial terms dominate. Dividing the numerator and denominator by \(x^4\) gives \(f(x)=\frac{100/x^2+e^x/x^4}{e^x/x^4-1}\). The numerator approaches \(0\) and the denominator approaches \(-1\), so \(\lim_{x\to-\infty}f(x)=0\).

Answer

\(\lim_{x\to\infty}f(x)=1\) and \(\lim_{x\to-\infty}f(x)=0\)
Infinite limits and asymptotes 29
Let \(g(x)=(x^2-5x)e^x\). Chiara claims, “As \(x\to-\infty\), the first factor approaches \(+\infty\) and the second factor approaches \(0\), so they cancel and the limit must be \(1\).” Evaluate Chiara's claim and find the correct limit as \(x\to-\infty\).

Hints

- A product with one factor approaching infinity and the other approaching zero does not have an automatic value. - Rewrite \(e^x\) using \(u=-x\). - Compare polynomial growth with exponential growth.

Solution

1. The claim is incorrect. A product of a quantity approaching infinity and a quantity approaching zero is an indeterminate form; the limit depends on their relative rates. 2. Let \(u=-x\). As \(x\to-\infty\), \(u\to\infty\), and \(g(x)=(u^2+5u)e^{-u}=\frac{u^2+5u}{e^u}\). 3. The exponential denominator grows faster than the polynomial numerator, so \(\frac{u^2+5u}{e^u}\to0\). Therefore, \(\lim_{x\to-\infty}g(x)=0\).

Answer

The claim is incorrect; \(\lim_{x\to-\infty}(x^2-5x)e^x=0\).
Infinite limits and asymptotes 30
Let \(f(x)=\frac{2x+4}{x-1}\). a) Find \(\lim_{x\to\infty}f(x)\). b) Find \(\lim_{x\to1^+}f(x)\) and \(\lim_{x\to1^-}f(x)\). c) State the horizontal and vertical asymptotes.

Hints

- Divide by \(x\) for the limit at infinity. - Determine the sign of the denominator on each side of \(x=1\). - Use the limits to identify the asymptotes.

Solution

1. Divide the numerator and denominator by \(x\): \(f(x)=\frac{2+\frac{4}{x}}{1-\frac{1}{x}}\). Therefore, \(f(x)\to2\) as \(x\to\infty\). 2. As \(x\to1^+\), the numerator approaches \(6\) and the denominator approaches \(0\) through positive values, so \(f(x)\to+\infty\). 3. As \(x\to1^-\), the numerator approaches \(6\) and the denominator approaches \(0\) through negative values, so \(f(x)\to-\infty\). 4. Thus, the horizontal asymptote is \(y=2\), and the vertical asymptote is \(x=1\).

Answer

a) \(2\) b) \(\lim_{x\to1^+}f(x)=+\infty\); \(\lim_{x\to1^-}f(x)=-\infty\) c) Horizontal: \(y=2\); vertical: \(x=1\)
Infinite limits and asymptotes 31
Evaluate each limit. a) \(\lim_{x\to\infty}\frac{x^2+4}{e^x}\) b) \(\lim_{x\to-\infty}(x-2)e^x\) c) \(\lim_{x\to0^+}\frac{3}{1-e^x}\)

Hints

- Compare polynomial and exponential growth rates. - Determine what \(e^x\) does as \(x\to-\infty\). - For a denominator approaching zero, determine whether it approaches from the positive or negative side.

Solution

1. In part a, \(e^x\) grows faster than the polynomial \(x^2+4\), so the quotient approaches \(0\). 2. In part b, let \(u=-x\). Then \((x-2)e^x=(-u-2)e^{-u}=-\frac{u+2}{e^u}\), which approaches \(0\) as \(u\to\infty\). 3. In part c, as \(x\to0^+\), \(e^x>1\), so \(1-e^x\to0^-\). Therefore, \(\frac{3}{1-e^x}\to-\infty\).

Answer

a) \(0\) b) \(0\) c) \(-\infty\)
Infinite limits and asymptotes 32
Let \(f(x)=\frac{x+2}{e^x-e^2}\). Determine the domain and analyze the behavior of \(f\) at every boundary of its domain.

Hints

- Find where the denominator is zero. - Compare linear and exponential growth for large positive \(x\). - Determine the limiting denominator as \(x\to-\infty\). - Analyze the signs on the two sides of the excluded value.

Solution

1. The denominator is zero when \(e^x=e^2\), so \(x=2\). Thus, the domain is \(D=\mathbb{R}\setminus\{2\}\). 2. As \(x\to\infty\), the exponential denominator grows faster than the linear numerator, so \(f(x)\to0\). 3. As \(x\to-\infty\), \(e^x\to0\), so the denominator approaches \(-e^2\), while the numerator approaches \(-\infty\). Therefore, \(f(x)\to\infty\). 4. As \(x\to2^+\), the numerator approaches \(4\) and the denominator approaches \(0^+\), so \(f(x)\to\infty\). As \(x\to2^-\), the denominator approaches \(0^-\), so \(f(x)\to-\infty\).

Answer

\(D=\mathbb{R}\setminus\{2\}\) \(\lim_{x\to\infty}f(x)=0\) \(\lim_{x\to-\infty}f(x)=\infty\) \(\lim_{x\to2^+}f(x)=\infty\) \(\lim_{x\to2^-}f(x)=-\infty\)
Infinite limits and asymptotes 33
Let \(g(x)=\frac{2x^2-\ln(x)}{x^2+5}\). Determine the behavior of \(g(x)\) as \(x\to\infty\).

Hints

- Divide the numerator and denominator by the highest power of \(x\). - Compare the growth of \(x^2\) with the growth of \(\ln(x)\). - Determine the limits of the smaller fractions separately. - Apply the quotient law for limits.

Solution

1. Divide the numerator and denominator by \(x^2\): \(g(x)=\frac{2-\frac{\ln(x)}{x^2}}{1+\frac{5}{x^2}}\). 2. As \(x\to\infty\), \(\frac{5}{x^2}\to0\). 3. A power function grows faster than a logarithmic function, so \(\frac{\ln(x)}{x^2}\to0\). 4. Therefore, \(\lim_{x\to\infty}g(x)=\frac{2-0}{1+0}=2\).

Answer

\(g(x)\to2\) as \(x\to\infty\).
Infinite limits and asymptotes 34
Let \(f(x)=\frac{\ln(x^2)+1}{x^2}\), where \(x\neq0\). Determine the behavior of \(f(x)\) as \(x\to0\) and as \(x\to\infty\).

Hints

- Examine the numerator and denominator separately as \(x\to0\). - Compare the growth of a logarithmic function with the growth of a power function. - Split the expression into two fractions for the limit at infinity.

Solution

1. As \(x\to0\), \(x^2\to0^+\) and \(\ln(x^2)+1\to-\infty\). Therefore, \(\frac{\ln(x^2)+1}{x^2}\to-\infty\). 2. As \(x\to\infty\), rewrite the function as \(f(x)=\frac{\ln(x^2)}{x^2}+\frac{1}{x^2}\). A power function grows faster than a logarithmic function, so \(\frac{\ln(x^2)}{x^2}\to0\), and \(\frac{1}{x^2}\to0\). 3. Thus, \(\lim_{x\to0}f(x)=-\infty\) and \(\lim_{x\to\infty}f(x)=0\).

Answer

As \(x\to0\), \(f(x)\to-\infty\). As \(x\to\infty\), \(f(x)\to0\).
Infinite limits and asymptotes 35
Let \(f(x)=\frac{1}{2}x^2+3x+4e^{-x/2}\). a) Show that the graph of \(f\) approaches a parabola \(q\) as \(x\to\infty\), and give the equation of \(q\). b) Determine whether the graph also approaches \(q\) as \(x\to-\infty\). Justify your answer with a limit.

Hints

- Identify the polynomial and exponential parts of the function. - Two graphs approach each other when the difference of their function values approaches \(0\). - Review the behavior of \(e^u\) as \(u\to\infty\) and \(u\to-\infty\). - Evaluate the two directions separately.

Solution

1. Separate the polynomial and exponential parts: \(q(x)=\frac{1}{2}x^2+3x\) and \(f(x)-q(x)=4e^{-x/2}\). 2. As \(x\to\infty\), \(\lim(f(x)-q(x))=\lim4e^{-x/2}=0\). Therefore, the graph of \(f\) approaches the parabola \(q(x)=\frac{1}{2}x^2+3x\). 3. As \(x\to-\infty\), the exponent \(-x/2\to\infty\), so \(\lim(f(x)-q(x))=\lim4e^{-x/2}=\infty\). Therefore, the graph does not approach \(q\) in that direction.

Answer

a) \(q(x)=\frac{1}{2}x^2+3x\), because \(\lim_{x\to\infty}(f(x)-q(x))=0\). b) No. Since \(\lim_{x\to-\infty}(f(x)-q(x))=\infty\), the graphs do not approach each other as \(x\to-\infty\).
Infinite limits and asymptotes 36
Let \(h(x)=\frac{e^x}{x+2}\), for \(x\ne-2\). a) Explain why \(h\) has no zeros. b) Find the limits of \(h(x)\) as \(x\to-2^-\), \(x\to-2^+\), \(x\to-\infty\), and \(x\to\infty\). c) Name every vertical or horizontal asymptote of the graph. Why is \(y=0\) an asymptote even though the graph never crosses it?

Hints

- Check whether the numerator can ever be zero. - Near the excluded input, compare the signs of the two factors. - For each end of the x-axis, consider the exponential and linear factors separately. - A horizontal asymptote describes end behavior, not an intersection.

Solution

1. The numerator \(e^x\) is always positive, so the quotient never equals \(0\). 2. Near \(-2\), the numerator approaches \(e^{-2}>0\). The denominator is negative from the left and positive from the right. Thus \(\lim_{x\to-2^-}h(x)=-\infty\) and \(\lim_{x\to-2^+}h(x)=+\infty\). 3. As \(x\to-\infty\), \(e^x\to0^+\) while \(x+2\to-\infty\), so \(h(x)\to0^-\). As \(x\to\infty\), exponential growth exceeds linear growth, so \(h(x)\to+\infty\). 4. Therefore \(x=-2\) is a vertical asymptote and \(y=0\) is a left-end horizontal asymptote. An asymptote concerns approaching a line, not crossing it.

Answer

a) \(e^x>0\) for every real \(x\), so \(h(x)\ne0\). b) Left of \(-2\): \(-\infty\); right of \(-2\): \(+\infty\); as \(x\to-\infty\): \(0\) from below; as \(x\to\infty\): \(+\infty\). c) Vertical asymptote \(x=-2\) and horizontal asymptote \(y=0\). The curve can approach the x-axis without reaching it.
Infinite limits and asymptotes 37
Let \(f(x)=\frac{\sqrt{x^2+1}}{x}\), \(x\ne0\). Find the one-sided limits as \(x\to0\), the limits as \(x\to\infty\) and \(x\to-\infty\), and state all vertical and horizontal asymptotes.

Hints

- Near \(0\), the numerator stays positive while the denominator changes sign. - At large magnitude, factor \(x^2\) from inside the square root carefully. - Remember that \(\sqrt{x^2}=|x|\), which behaves differently for positive and negative \(x\).

Solution

1. As \(x\to0\), the numerator approaches \(1\). The denominator approaches \(0^-\) from the left and \(0^+\) from the right, so \(\lim_{x\to0^-}f(x)=-\infty\) and \(\lim_{x\to0^+}f(x)=+\infty\). Thus \(x=0\) is a vertical asymptote. 2. For \(x>0\), \(f(x)=\sqrt{1+\frac{1}{x^2}}\), so \(\lim_{x\to\infty}f(x)=1\). 3. For \(x<0\), \(\sqrt{x^2}=|x|=-x\), so \(f(x)=-\sqrt{1+\frac{1}{x^2}}\). Hence \(\lim_{x\to-\infty}f(x)=-1\). The horizontal asymptotes are \(y=1\) to the right and \(y=-1\) to the left.

Answer

\(\lim_{x\to0^-}f(x)=-\infty\), \(\lim_{x\to0^+}f(x)=+\infty\). \(\lim_{x\to\infty}f(x)=1\), \(\lim_{x\to-\infty}f(x)=-1\). Vertical asymptote: \(x=0\). Horizontal asymptotes: \(y=1\) as \(x\to\infty\), and \(y=-1\) as \(x\to-\infty\).
Infinite limits and asymptotes 38
Let \(g(x)=\frac{1}{1-e^{-x}}\), \(x\ne0\). Determine the behavior as \(x\to0^-\), \(x\to0^+\), \(x\to\infty\), and \(x\to-\infty\). State the vertical asymptote and any horizontal asymptotes.

Hints

- Near \(0\), determine whether \(e^{-x}\) is slightly greater than or less than \(1\) on each side. - At positive infinity, track the exponential decay of \(e^{-x}\). - At negative infinity, recognize that \(e^{-x}\) grows without bound.

Solution

1. As \(x\to0^-\), \(e^{-x}>1\) and approaches \(1\), so \(1-e^{-x}\to0^-\) and \(g(x)\to-\infty\). As \(x\to0^+\), the denominator approaches \(0^+\), so \(g(x)\to+\infty\). Thus \(x=0\) is a vertical asymptote. 2. As \(x\to\infty\), \(e^{-x}\to0\), so \(g(x)\to1\). 3. As \(x\to-\infty\), \(e^{-x}\to+\infty\), so the denominator tends to \(-\infty\) and \(g(x)\to0\). Therefore the right-end horizontal asymptote is \(y=1\), while the left-end horizontal asymptote is \(y=0\).

Answer

\(\lim_{x\to0^-}g(x)=-\infty\), \(\lim_{x\to0^+}g(x)=+\infty\). \(\lim_{x\to\infty}g(x)=1\), \(\lim_{x\to-\infty}g(x)=0\). Vertical asymptote: \(x=0\). Horizontal asymptotes: \(y=1\) as \(x\to\infty\), and \(y=0\) as \(x\to-\infty\).
Infinite limits and asymptotes 39
Let \(f(x)=\frac{x^2+1}{x-1}\). Find the vertical asymptote and the slant asymptote. Justify the slant asymptote with a limit involving the difference between \(f(x)\) and a line.

Hints

- Divide the numerator by the denominator to separate a line from a remainder term. - An asymptotic line is confirmed when the difference between the function and the line approaches zero. - Check excluded denominator zeros separately for vertical behavior.

Solution

1. Polynomial division gives \(f(x)=x+1+\frac{2}{x-1}\). 2. As \(x\to\pm\infty\), \(\frac{2}{x-1}\to0\), so \(f(x)-(x+1)\to0\). Thus the slant asymptote is \(y=x+1\). 3. At \(x=1\), the numerator approaches \(2\) while the denominator approaches \(0\), giving unbounded one-sided behavior. Thus \(x=1\) is a vertical asymptote.

Answer

Vertical asymptote: \(x=1\). Slant asymptote: \(y=x+1\).
Infinite limits and asymptotes 40
Let \(f(x)=\frac{3x}{\sqrt{x^2+9}}\). Find the horizontal asymptote approached as \(x\to\infty\) and the horizontal asymptote approached as \(x\to-\infty\). Explain why the two limits have opposite signs.

Hints

- Factor the largest power of \(x\) from inside the square root. - Remember that \(\sqrt{x^2}=|x|\), especially when \(x\) is negative.

Solution

1. Write \(\sqrt{x^2+9}=|x|\sqrt{1+9/x^2}\). 2. As \(x\to\infty\), \(|x|=x\), so \(f(x)=\frac{3}{\sqrt{1+9/x^2}}\to3\). 3. As \(x\to-\infty\), \(|x|=-x\), so \(f(x)=\frac{-3}{\sqrt{1+9/x^2}}\to-3\). 4. The sign difference comes from \(x/|x|\), which approaches \(1\) on the right and \(-1\) on the left.

Answer

As \(x\to\infty\), \(f(x)\to3\), so \(y=3\) is the right-end horizontal asymptote. As \(x\to-\infty\), \(f(x)\to-3\), so \(y=-3\) is the left-end horizontal asymptote.
Infinite limits and asymptotes 41
Bongani says, “Once a graph has a horizontal asymptote, it can never cross that line.” Consider \(f(x)=1+\frac{\sin x}{x}\), for \(x\ne0\). Determine the horizontal asymptote as \(x\to\pm\infty\), and use this function to evaluate Bongani's claim.

Hints

- Separate the question of long-run behavior from what can happen at finite x-values. - Look for inputs where the added fraction becomes exactly zero.

Solution

1. Since \(-1\le\sin x\le1\), the quotient \(\frac{\sin x}{x}\to0\) as \(x\to\infty\) and as \(x\to-\infty\). 2. Therefore \(f(x)\to1\) on both ends, so \(y=1\) is a horizontal asymptote. 3. Whenever \(x=n\pi\) for a nonzero integer \(n\), \(\sin x=0\), so \(f(n\pi)=1\). 4. Thus the graph meets the horizontal asymptote infinitely many times. A horizontal asymptote describes end behavior; it does not forbid crossings at finite inputs.

Answer

The horizontal asymptote is \(y=1\) as \(x\to\pm\infty\). Bongani's claim is false because \(f(n\pi)=1\) for every nonzero integer \(n\), so the graph crosses or meets its horizontal asymptote repeatedly.
Infinite limits and asymptotes 42
Use the graph of \(s\) on \([0,2\pi]\) to locate the vertical asymptotes. Describe the one-sided behavior at each asymptote.
Figure for problem 542625

Hints

- Look for labeled x-values where neighboring branches leave the visible vertical scale. - Read the direction of each branch separately from the left and from the right. - Match each unbounded branch pair to the exact \(\pi\)-based tick label beneath it.

Solution

1. The graph becomes unbounded at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\), so these are the vertical asymptotes. 2. At \(x=\frac{\pi}{2}\), the left branch increases without bound and the right branch decreases without bound. Thus the left-hand limit is \(+\infty\) and the right-hand limit is \(-\infty\). 3. At \(x=\frac{3\pi}{2}\), the left branch decreases without bound and the right branch increases without bound. Thus the left-hand limit is \(-\infty\) and the right-hand limit is \(+\infty\).

Answer

Vertical asymptotes: \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). At \(x=\frac{\pi}{2}\): left limit \(+\infty\), right limit \(-\infty\). At \(x=\frac{3\pi}{2}\): left limit \(-\infty\), right limit \(+\infty\).
Infinite limits and asymptotes 43
Consider \(r(x)=\sqrt{x^2+9}\). The graph has no vertical asymptotes, but it approaches a different line at each end. Find the line approached as \(x\to\infty\) and the line approached as \(x\to-\infty\).

Hints

- To show a line is an asymptote, examine the difference between the function and that line. - The square root behaves like \(|x|\) for large \(|x|\), so the two ends should be treated separately.

Solution

1. For \(x\to\infty\), compare \(r(x)\) with \(x\). Rationalizing gives \(r(x)-x=\frac{9}{\sqrt{x^2+9}+x}\to0\). Thus the right-end slant asymptote is \(y=x\). 2. For \(x\to-\infty\), compare \(r(x)\) with \(-x\). Then \(r(x)-(-x)=\sqrt{x^2+9}+x\). 3. Rationalizing this difference gives \(\sqrt{x^2+9}+x=\frac{9}{\sqrt{x^2+9}-x}\to0\) as \(x\to-\infty\). 4. Therefore the left-end slant asymptote is \(y=-x\).

Answer

As \(x\to\infty\), the graph approaches \(y=x\). As \(x\to-\infty\), the graph approaches \(y=-x\).
Infinite limits and asymptotes 44
Analyze \(F(x)=\frac{e^x}{e^x-2}\). Find its vertical asymptote with the two one-sided infinite limits, and find the horizontal asymptote approached at each end of the x-axis.

Hints

- Solve for the input where the denominator vanishes, then check the denominator's sign on each side. - For positive-end behavior, divide through by the dominant exponential term. - For negative-end behavior, use the fact that \(e^x\) approaches zero.

Solution

1. The denominator is zero when \(e^x=2\), so the possible vertical asymptote is \(x=\ln2\). 2. As \(x\to(\ln2)^-\), the denominator approaches \(0^-\) while the numerator approaches \(2\), so \(F(x)\to-\infty\). From the right, the denominator approaches \(0^+\), so \(F(x)\to+\infty\). 3. As \(x\to\infty\), divide numerator and denominator by \(e^x\): \(F(x)=\frac{1}{1-2e^{-x}}\to1\). 4. As \(x\to-\infty\), \(e^x\to0\), so \(F(x)\to0\).

Answer

Vertical asymptote: \(x=\ln2\), with left limit \(-\infty\) and right limit \(+\infty\). Horizontal asymptotes: \(y=1\) as \(x\to\infty\) and \(y=0\) as \(x\to-\infty\).
Infinite limits and asymptotes 45
Let \(f(x)=e^{1/x}\), for \(x\ne0\). Find the two one-sided limits as \(x\to0\) and the limits as \(x\to\pm\infty\). Use them to identify every vertical and horizontal asymptote.

Hints

- First determine what \(1/x\) does from each side of zero. - A vertical asymptote does not require both one-sided limits to be infinite. - At very large \(|x|\), the exponent itself approaches a simple value.

Solution

1. As \(x\to0^+\), \(1/x\to+\infty\), so \(e^{1/x}\to+\infty\). This is enough to make \(x=0\) a vertical asymptote. 2. As \(x\to0^-\), \(1/x\to-\infty\), so \(e^{1/x}\to0\). The two sides therefore have very different behavior. 3. As \(x\to\infty\) or \(x\to-\infty\), \(1/x\to0\), so \(e^{1/x}\to e^0=1\). 4. Thus \(y=1\) is a horizontal asymptote on both ends.

Answer

\(\lim_{x\to0^+}f(x)=+\infty\) and \(\lim_{x\to0^-}f(x)=0\), so \(x=0\) is a vertical asymptote. \(\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=1\), so \(y=1\) is the horizontal asymptote on both ends.
Infinite limits and asymptotes 46
Consider \(f(x)=\sqrt{x^2+x}-x\) on its real domain. Find \(\lim_{x\to\infty}f(x)\) by rationalizing, and state the corresponding horizontal asymptote. Also describe what happens as \(x\to-\infty\).

Hints

- A difference of two large quantities suggests multiplying by a conjugate. - Treat positive and negative large x-values separately because \(\sqrt{x^2}=|x|\). - A finite end limit gives a horizontal asymptote on that end only.

Solution

1. For \(x\ne0\), rationalize: \(\sqrt{x^2+x}-x=\frac{x}{\sqrt{x^2+x}+x}\). This identity is sufficient for both end limits. 2. For \(x\to\infty\), divide numerator and denominator by \(x>0\) to get \(\frac{1}{\sqrt{1+1/x}+1}\to\frac{1}{2}\). Thus \(y=\frac{1}{2}\) is a right-end horizontal asymptote. 3. As \(x\to-\infty\), \(\sqrt{x^2+x}\sim -x\), so \(\sqrt{x^2+x}-x\sim-2x\to+\infty\). There is no horizontal asymptote on the left.

Answer

\(\lim_{x\to\infty}f(x)=\frac{1}{2}\), so the right-end horizontal asymptote is \(y=\frac{1}{2}\). As \(x\to-\infty\), \(f(x)\to+\infty\).

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