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Chapter 5: Q. 70 (page 479)

The Limit Comparison Test: If two functions f(x) and g(x) behave in a similar way on 摆补,鈭�), then their improper integrals on 摆补,鈭�) should converge or diverge together. More specifically, the limit comparison test says that if $\lim_{x 鈫� 鈭�} \frac{f (x)}{g (x)} = L$ for some positive real number L, then the improper integrals ${鈭�}_{a}^{鈭�} f (x) d x and {鈭�}_{a}^{鈭�} g (x) d x$ either both converge or both diverge. We will learn more about limit comparison tests in Chapter 7. Use this test to determine the convergence or divergence of the integrals in Exercises 65鈥�70.
${鈭�}_{1}^{鈭�} \frac{1 + e^{x}}{x e^{x}} d x$

Short Answer

Expert verified

The given integral is diverges.

Step by step solution

01

Step 1. Given Information.

The given integral is ${鈭�}_{1}^{鈭�} \frac{1 + e^{x}}{x e^{x}} d x .$

02

Step 2. Determining the integral is convergence or divergence.

To determine the convergence or divergence of the integral, we will use the limit comparison test.

Let two functionsf(x)andg(x)as: $f (x) = \frac{1}{x} and g (x) = \frac{1 + e^{x}}{x e^{x}} .$

03

Step 3. Estimate the limits of the functions.

So,

$\lim_{x 鈫� 鈭�} \frac{f (x)}{g (x)} = \lim_{x 鈫� 鈭�} \frac{(\frac{1}{x})}{(\frac{1 + e^{x}}{x e^{x}})} = \lim_{x 鈫� 鈭�} \frac{x e^{x}}{x (1 + e^{x})} = \lim_{x 鈫� 鈭�} \frac{e^{x}}{(1 + e^{x})} = \lim_{x 鈫� 鈭�} \frac{1 + e^{x} - 1}{(1 + e^{x})} = \lim_{x 鈫� 鈭�} 1 - \lim_{x 鈫� 鈭�} \frac{1}{(1 + e^{x})} = 1 [\frac{1}{鈭�} = 0]$

Since 1 is a positive number. Thus, by the limit comparison test, both the improper integrals ${鈭�}_{1}^{鈭�} \frac{1}{x} d x and {鈭�}_{1}^{鈭�} \frac{1 + e^{x}}{x e^{x}} d x$ either both converge or diverge.

By improper integral for power functions ${鈭�}_{1}^{鈭�} \frac{1}{x} d x$ diverges.

Thus, the improper integral ${鈭�}_{1}^{鈭�} \frac{1 + e^{x}}{x e^{x}} d x$ also diverges.

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