Free Math Worksheets

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

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 \(x\).

The limit is \(2\), so \(y=2\) is a horizontal asymptote.

- 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 \(x=1\).

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 \(x=1\).

\(\lim_{x\to 1}f(x)\) 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.

- 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?

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, \(\lim_{x\to 5}f_n(x)=+\infty\).

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, \(\lim_{x\to 5}f_n(x)=+\infty\).

- 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.

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.

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.

- 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.

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 \(\lim_{x\to -3^+}f(x)=-\infty\).

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 \(\lim_{x\to -3^+}f(x)=-\infty\).

- 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 \(5\).

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, \(\lim_{x\to 5^+}h(x)=-\infty\).

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 \(\lim_{x\to 5^+}h(x)=-\infty\).

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

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 \(+\infty\).

a) \(-\infty\) b) \(+\infty\)

- 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.

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 \(x=1\).

a) \(2\) b) \(\lim_{x\to1^+}f(x)=+\infty\); \(\lim_{x\to1^-}f(x)=-\infty\) c) Horizontal: \(y=2\); vertical: \(x=1\)

- 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.

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 \(f(x)\to-\infty\).

\(D=\mathbb{R}\setminus\{2\}\) \(\lim_{x\to\infty}f(x)=0\) \(\lim_{x\to-\infty}f(x)=\infty\) \(\lim_{x\to2^+}f(x)=\infty\) \(\lim_{x\to2^-}f(x)=-\infty\)

- Require the expression under the square root to be nonnegative. - Use the sign before the radical to determine the range. - Apply the chain rule to find the derivative. - Use the point and derivative value to write the tangent line. - Interpret an unbounded derivative near an endpoint geometrically.

1. a) Require \(2x+6\ge0\), so \(D_f=[-3,\infty)\). Since \(\sqrt{2x+6}\ge0\) and grows without bound, the range is \((-\infty,4]\). 2. b) Differentiate: \(f'(x)=-\frac{1}{\sqrt{2x+6}}\). 3. At \(x=\frac{3}{2}\), \(f\left(\frac{3}{2}\right)=1\) and \(f'\left(\frac{3}{2}\right)=-\frac{1}{3}\). 4. Using point-slope form, \(y-1=-\frac{1}{3}\left(x-\frac{3}{2}\right)\), so \(t:y=-\frac{1}{3}x+\frac{3}{2}\). 5. c) As \(x\to-3^+\), \(\sqrt{2x+6}\to0^+\), so \(f'(x)\to-\infty\). The graph has a vertical tangent at its endpoint \((-3,4)\).

a) \(D_f=[-3,\infty)\); range: \((-\infty,4]\) b) \(t:y=-\frac{1}{3}x+\frac{3}{2}\) c) \(\lim_{x\to-3^+}f'(x)=-\infty\); the graph has a vertical tangent at \((-3,4)\).

- Require the square-root radicand to be nonnegative. - Replace \(x\) with \(-x\) to test symmetry. - Apply the chain rule to differentiate. - Analyze the signs of the numerator and denominator near each endpoint. - An unbounded derivative magnitude indicates a vertical tangent.

1. a) Require \(9-x^2\ge0\), so \(D_f=[-3,3]\). Since \(f(-x)=f(x)\), the graph is symmetric about the y-axis. It is not symmetric about the x-axis because it contains only nonnegative y-values. 2. b) By the chain rule, \(f'(x)=-\frac{2x}{\sqrt{9-x^2}}\). 3. As \(x\to3^-\), the numerator approaches \(-6\) and the denominator approaches \(0^+\), so \(f'(x)\to-\infty\). 4. As \(x\to-3^+\), the numerator approaches \(6\) and the denominator approaches \(0^+\), so \(f'(x)\to\infty\). 5. Therefore, the graph has vertical tangent lines at \((-3,0)\) and \((3,0)\).

a) \(D_f=[-3,3]\); the graph is symmetric about the y-axis and not symmetric about the x-axis. b) \(f'(x)=-\frac{2x}{\sqrt{9-x^2}}\); \(f'(x)\to-\infty\) as \(x\to3^-\), and \(f'(x)\to\infty\) as \(x\to-3^+\). Both endpoint tangents are vertical.

- Set the radicand greater than or equal to \(0\) to find the domain. - Use the derivative and graph point to write the tangent line. - Examine the radical denominator as \(x\) approaches each endpoint. - A derivative that becomes unbounded does not have a finite endpoint value.

1. a) Require \(100-4x^2\ge0\), so \(x^2\le25\) and the domain is \([-5,5]\). The x-intercepts are \((-5,0)\) and \((5,0)\). 2. Differentiate: \(g'(x)=-\frac{4x}{\sqrt{100-4x^2}}\). 3. b) At \(x=3\), \(g(3)=8\) and \(g'(3)=-\frac{3}{2}\). Thus, \(y-8=-\frac{3}{2}(x-3)\), or \(y=-\frac{3}{2}x+\frac{25}{2}\). 4. c) As \(x\to5^-\), \(g'(x)\to-\infty\). As \(x\to-5^+\), \(g'(x)\to\infty\). 5. The endpoint slopes do not approach finite real values, so the function is not differentiable at the endpoints; the graph has vertical tangents there.

a) Domain \([-5,5]\); x-intercepts \((-5,0)\) and \((5,0)\) b) \(y=-\frac{3}{2}x+\frac{25}{2}\) c) The slope approaches \(-\infty\) at \(5\) from the left and \(\infty\) at \(-5\) from the right, so \(g\) is not differentiable at either endpoint.

- 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.

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.

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 \(x\to-\infty\).

- Find the vertical asymptote and compare the one-sided behavior. - Determine the sign of \(f'\) and relate it to the monotonicity of \(f\). - Evaluate the end behavior of each formula as \(x \to \pm\infty\).

1. The function is undefined at \(x = 1\). For \(f\), the two sides of the vertical asymptote have opposite signs: \(f(x) \to -\infty\) as \(x \to 1^-\) and \(f(x) \to \infty\) as \(x \to 1^+\). This matches Figure 1. 2. Differentiate: \(f'(x) = -\frac{2}{(x-1)^2}\). This derivative is negative for every \(x \ne 1\), so \(f\) is decreasing on each part of its domain. Near \(x = 1\), the derivative approaches \(-\infty\) from both sides, matching Figure 2. 3. As \(x \to \pm\infty\), \(\frac{2}{x-1} \to 0\), so \(f(x) \to -1\). The horizontal asymptote of \(f\) is \(y = -1\). 4. As \(x \to \pm\infty\), \(-\frac{2}{(x-1)^2} \to 0\), so the horizontal asymptote of \(f'\) is \(y = 0\).

a) Figure 1 shows \(f\), and Figure 2 shows \(f'\). b) For \(f\): \(y = -1\) For \(f'\): \(y = 0\)

- Examine the behavior near the excluded input and for large \(|x|\). - Set the first derivative equal to zero on the positive branch. - Perpendicular nonvertical lines have negative-reciprocal slopes. - Use point-slope form for the normal line.

1. The function is undefined at \(x=0\), giving the vertical asymptote \(x=0\). Since \(\frac{2}{x}\to0\) as \(|x|\to\infty\), the slant asymptote is \(y=\frac{1}{2}x\). 2. Differentiate: \(f'(x)=\frac{1}{2}-\frac{2}{x^2}\). Setting this equal to zero gives \(x^2=4\). For \(x>0\), \(x=2\). Since \(f''(2)=\frac{4}{2^3}=\frac{1}{2}>0\), the point is a local minimum. Its coordinates are \((2,2)\). 3. At \(x=1\), the point is \(\left(1,\frac{5}{2}\right)\), and the tangent slope is \(-\frac{3}{2}\). The normal slope is \(\frac{2}{3}\). Thus \(y-\frac{5}{2}=\frac{2}{3}(x-1)\), or \(y=\frac{2}{3}x+\frac{11}{6}\).

a) \(x=0\) and \(y=\frac{1}{2}x\). b) \((2,2)\). c) \(y=\frac{2}{3}x+\frac{11}{6}\).

- 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?

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 \(y=1\).

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) \(y=1\)