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Chapter 4: Problem 29

Evaluate the following limits. $$\lim _{x \rightarrow \infty} \frac{e^{1 / x}-1}{1 / x}$$

Short Answer

Expert verified
Answer: The limit of the function as x approaches infinity is 1.

Step by step solution

01

Recognize the indeterminate form

First, let's see if the limit has an indeterminate form (in this case, it would be 0/0). As x approaches infinity: $$\lim_{x \rightarrow \infty} e^{1/x} \rightarrow e^{0} = 1$$ So, we have: $$\lim_{x \rightarrow \infty} \frac{e^{1/x} - 1}{1/x} = \frac{1-1}{0} = \frac{0}{0}$$ Since this function has an indeterminate form, we need to use L'Hopital's rule.
02

Apply L'Hopital's Rule

Now that we've identified this as an indeterminate form, we can apply L'Hopital's Rule. We will differentiate the numerator and denominator separately. The derivative of the numerator with respect to x: $$\frac{d}{dx}(e^{1/x} - 1) = -\frac{1}{x^2}e^{1/x}$$ The derivative of the denominator with respect to x: $$\frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}$$ Now we replace the original limit by the limit of their derivatives: $$\lim_{x \rightarrow \infty} \frac{-\frac{1}{x^2}e^{1/x}}{-\frac{1}{x^2}}$$
03

Simplify and evaluate the limit

Now, we need to simplify the expression and evaluate the limit. To do this, we can cancel out the \(-\frac{1}{x^2}\) terms in the numerator and the denominator: $$\lim_{x \rightarrow \infty} e^{1/x}$$ Now, as x approaches infinity, the limit becomes: $$\lim_{x \rightarrow \infty} e^{1/x} = e^0 = 1$$ So the limit of the given function as x approaches infinity is 1.

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