Free Math Worksheets

- Divide by the highest power of \(x\). - Compare the leading coefficients.

1. Divide the numerator and denominator by \(x\): \(k(x)=\frac{10+\frac{5}{x}}{2-\frac{4}{x}}\). 2. As \(x\to\infty\) or \(x\to-\infty\), the reciprocal terms approach \(0\). 3. Therefore, \(k(x)\to\frac{10}{2}=5\) in both directions.

\(\lim_{x\to\infty}k(x)=5\) and \(\lim_{x\to-\infty}k(x)=5\)

- Divide by the highest power of \(x\) in the denominator. - Compare the degrees of the numerator and denominator.

1. For part a, divide by \(x^2\): \(f(x)=\frac{6+\frac{4}{x}}{3-\frac{1}{x^2}}\). The reciprocal terms approach \(0\), so the limit is \(\frac{6}{3}=2\) in both directions. 2. For part b, the numerator remains constant while the denominator grows without bound as \(|x|\to\infty\). Therefore, the quotient approaches \(0\) in both directions.

a) \(\lim_{x\to\pm\infty}f(x)=2\) b) \(\lim_{x\to\pm\infty}g(x)=0\)

- Divide by or factor out the dominant power of \(x\). - Simplify algebraically before taking the limit whenever possible.

1. Divide by \(x\): \(\frac{7-\frac{2}{x}}{\frac{3}{x}-2}\to\frac{7}{-2}=-\frac{7}{2}\). 2. The denominator grows without bound, so the quotient approaches \(0\). 3. Factor \(x^3\): \(x^2-x^3=x^3\left(\frac{1}{x}-1\right)\). The second factor approaches \(-1\), so the expression approaches \(-\infty\). 4. Simplify: \(\frac{x^2+1}{x}-x=x+\frac{1}{x}-x=\frac{1}{x}\), which approaches \(0\).

1) \(-\frac{7}{2}\) 2) \(0\) 3) \(-\infty\) 4) \(0\)

- What happens when you substitute \(x=3\) directly? - Factor the numerator and denominator. - Is there a common factor that can be canceled? - What does a finite limit imply about the type of discontinuity?

1. Direct substitution gives the indeterminate form \(\frac{0}{0}\). 2. Factor the numerator: \(x^2-9=(x-3)(x+3)\). 3. Factor the denominator: \(2x^2-6x=2x(x-3)\). 4. For \(x\neq 3\), cancel the common factor: \(f(x)=\frac{x+3}{2x}\). 5. Evaluate the limit: \(\lim_{x\to 3}f(x)=\frac{3+3}{2(3)}=1\). 6. Because the limit exists and is finite, \(x=3\) is a removable discontinuity.

\(\lim_{x\to 3}f(x)=1\). The discontinuity at \(x=3\) is removable.

- 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 \(x\).

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

\(\lim_{x\to\infty}f(x)=0\) and \(\lim_{x\to-\infty}f(x)=-\infty\)

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

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, \(g(x)\to\infty\).

\(\lim_{x\to\infty}g(x)=-\infty\) and \(\lim_{x\to-\infty}g(x)=\infty\)

- 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.

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 \(\lim_{x\to-\infty}f(x)=0\).

\(\lim_{x\to\infty}f(x)=1\) and \(\lim_{x\to-\infty}f(x)=0\)

- 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.

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, \(\lim_{x\to-\infty}g(x)=0\).

The claim is incorrect; \(\lim_{x\to-\infty}(x^2-5x)e^x=0\).

- Factor each numerator completely. - Cancel only after noting that the original input is excluded. - Substitute into the simplified expression.

1. Factor: \(x^2-16=(x-4)(x+4)\) and \(2x-8=2(x-4)\). For \(x\ne4\), the expression equals \(\frac{x+4}{2}\). Its limit at \(4\) is \(4\). 2. Factor: \(x^2+4x+3=(x+1)(x+3)\). For \(x\ne-1\), the expression equals \(x+3\). Its limit at \(-1\) is \(2\).

1) \(4\) 2) \(2\)

- What must be true of the numerator for the discontinuity to be removable? - Substitute the excluded input into the numerator and determine the required output. - After finding the parameter, factor and simplify before evaluating the limit.

1. For the discontinuity at \(x=4\) to be removable, the numerator must also be zero at \(x=4\). 2. Substitute \(x=4\) into the numerator: \(2(4)^2-4k=0\). 3. Solve: \(32-4k=0\), so \(k=8\). 4. With \(k=8\), \(g(x)=\frac{2x^2-8x}{x-4}=\frac{2x(x-4)}{x-4}\). 5. For \(x\neq 4\), the expression simplifies to \(g(x)=2x\). 6. Therefore, \(\lim_{x\to 4}g(x)=\lim_{x\to 4}2x=8\).

\(k=8\), and \(\lim_{x\to 4}g(x)=8\).

- What condition must the denominator satisfy for \(x=2\) to be excluded from the domain? - What must also be true of the numerator for the discontinuity to be removable? - Use the denominator condition to form an equation for \(a\). - After finding \(a\), factor and simplify before evaluating the limit.

1. For \(x=2\) to be a discontinuity, the denominator must equal zero there: \(2^2+2a-6=0\). 2. Solve \(4+2a-6=0\), which gives \(a=1\). 3. With \(a=1\), factor the denominator: \(x^2+x-6=(x-2)(x+3)\). Factor the numerator: \(x^3-4x=x(x-2)(x+2)\). 4. For \(x\neq 2\), simplify: \(g(x)=\frac{x(x-2)(x+2)}{(x-2)(x+3)}=\frac{x(x+2)}{x+3}\). The factor causing the discontinuity cancels completely, so the discontinuity is removable. 5. Evaluate the limit: \(\lim_{x\to 2}g(x)=\frac{2(2+2)}{2+3}=\frac{8}{5}=1.6\).

\(a=1\), and \(\lim_{x\to 2}g(x)=\frac{8}{5}=1.6\).

- Find the limit in terms of \(a\) using the leading coefficients. - Set that expression equal to the required limit.

1. Divide the numerator and denominator by \(x^2\): \(g(x)=\frac{a+\frac{5}{x}}{3-\frac{2}{x^2}}\). 2. Therefore, \(\lim_{x\to\infty}g(x)=\frac{a}{3}\). 3. Set \(\frac{a}{3}=4\), which gives \(a=12\).

\(a=12\)

- Use the ratio of leading coefficients for part a. - Set the limit equal to the desired asymptote value. - Factor and cancel in part c while retaining the domain restriction.

1. Divide the numerator and denominator of \(f_a\) by \(x\): \(f_a(x)=\frac{a+\frac{6}{x}}{2-\frac{4}{x}}\). Thus, \(L=\frac{a}{2}\). 2. A horizontal asymptote \(y=5\) requires \(\frac{a}{2}=5\), so \(a=10\). 3. For \(x\ne0\), \(g(x)=\frac{x(2x+3)}{x}=2x+3=h(x)\). Therefore, both functions approach \(+\infty\) as \(x\to\infty\), and they agree everywhere that \(g\) is defined.

a) \(L=\frac{a}{2}\) b) \(a=10\) c) Both approach \(+\infty\), and \(g(x)=h(x)\) for \(x\ne0\).

- When can a denominator factor be canceled? - What must be true of the numerator at \(x=5\) for the discontinuity to be removable? - What kind of graph results when the simplified expression is linear?

1. For the discontinuity at \(x=5\) to be removable, the numerator must also equal zero at \(x=5\). 2. Solve \(5^2-5k-10=0\): \(15-5k=0\), so \(k=3\). 3. For \(k=3\), factor the numerator: \(x^2-3x-10=(x-5)(x+2)\). 4. For \(x\neq 5\), simplify: \(f(x)=\frac{(x-5)(x+2)}{x-5}=x+2\). 5. Therefore, \(\lim_{x\to 5}f(x)=5+2=7\). 6. The graph is the line \(y=x+2\) with an open point at \((5, 7)\), because \(x=5\) remains excluded from the original domain.

a) \(k=3\) b) \(\lim_{x\to 5}f(x)=7\) c) The graph is the line \(y=x+2\) with an open point at \((5, 7)\).

- 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.

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, \(\frac{3}{1-e^x}\to-\infty\).

a) \(0\) b) \(0\) c) \(-\infty\)

- Track the behavior of both factors as \(x\to0^+\). - Expand the polynomial factor into two terms. - Recall the standard limit \(x^p\ln(x)\to0\) for \(p>0\). - Apply the sum law for limits.

1. Expand the product: \(f(x)=x^2\ln(x)+3x\ln(x)\). 2. For every positive exponent \(p\), \(x^p\ln(x)\to0\) as \(x\to0^+\). The power factor approaches \(0\) rapidly enough to dominate the logarithm's divergence to \(-\infty\). 3. Therefore, \(x^2\ln(x)\to0\) and \(3x\ln(x)\to0\). 4. Hence, \(\lim_{x\to0^+}(x^2+3x)\ln(x)=0\).

\(\lim_{x\to0^+}(x^2+3x)\ln(x)=0\)

- 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.

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, \(\lim_{x\to\infty}g(x)=\frac{2-0}{1+0}=2\).

\(g(x)\to2\) as \(x\to\infty\).

- 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.

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 \(\lim_{x\to\infty}f(x)=0\).

As \(x\to0\), \(f(x)\to-\infty\). As \(x\to\infty\), \(f(x)\to0\).

- Find the zero of each function from its linear factor. - Compare the end behavior as \(x\to\infty\). - Check whether a graph approaches the x-axis from above or below. - Use the y-intercepts to distinguish graphs when needed.

1. For \(f(x)=(x+2)e^{-x}\), the only zero is \(x=-2\). Also, \(f(x)\to0^+\) as \(x\to\infty\). These features match Graph A. 2. For \(g(x)=(x-2)e^x\), the zero is \(x=2\). As \(x\to\infty\), both factors are positive and grow, so \(g(x)\to\infty\). These features match Graph B. 3. For \(h(x)=(2-x)e^{-x}\), the zero is \(x=2\). For \(x>2\), the function is negative, and \(h(x)\to0^-\) as \(x\to\infty\). These features match Graph C. 4. Graph D is the unused graph; it represents \((x+2)e^x\).

1. Graph A 2. Graph B 3. Graph C

- Distinguish between crossing and touching the x-axis. - Check the location and multiplicity of each zero. - Evaluate each option at \(x=0\). - Compare the graph's end behavior with \(e^x\) and \(e^{-x}\).

1. The graph touches the x-axis at \(x=2\), indicating a zero of even multiplicity. This eliminates (1), which has a simple zero, and (3), whose zero is at \(x=-2\). 2. The graph grows without bound as \(x\to\infty\). Choice (4) would approach \(0\) because of the factor \(e^{-x}\), so (4) is eliminated. 3. Choice (2), \(f(x)=(x-2)^2e^x\), also gives \(f(0)=4\), matching the y-intercept. It approaches \(0\) as \(x\to-\infty\), which matches the graph.

(2) \(f(x)=(x-2)^2e^x\)