Limit of x(e^(1/x)-1) as x approaches infinity: Solving with L'Hopital's Rule

[James Brady]

Homework Statement

Find $\lim_{x\to\infty} x(e^{1/x}-1)$

Homework Equations

$\lim_{x\to\infty} \frac{f(x)}{g(x)} = \lim_{x\to\infty} \frac{f'(x)}{g'(x)}$

The Attempt at a Solution

I attempted to rewrite the function in terms of a ratio and then use L'Hopital's rule:

$\lim_{x\to\infty} \frac{x}{(e^{1/x}-1)^{-1}} = \lim_{x\to\infty} \frac{1}{-(e^{1/x}-1)^{-2}(\frac{1}{x}e^{1/x})}$

The problem is that the exponential terms never go away. The bigger problem is I believe L'Hopital's Rule is probably unnecessary and I'm missing something more basic here.

 

[BvU]

Would it be easier for you to investigate $$\lim_{\varepsilon\downarrow 0} {e^\varepsilon -1 \over \varepsilon } \ \ \rm ? $$

 

[PeroK]

Another idea is to use power series.

 

[James Brady]

Would it be easier for you to investigate $$\lim_{\varepsilon\downarrow 0} {e^\varepsilon -1 \over \varepsilon } \ \ \rm ? $$

I am able to see what you did today . $\epsilon = 1/x , x = 1/ \epsilon$ Now I can infer that as x goes to infinity, epsilon goes to zero, so by making this replacement, we also replace the limit from infinity to zero. Then once we get into this nice form, L'Hopital's Rule and we're good. I'm not sure if I've ever seen this technique before, at least I don't remember it.

 

[vela]

I attempted to rewrite the function in terms of a ratio and then use L'Hopital's rule:

$\lim_{x\to\infty} \frac{x}{(e^{1/x}-1)^{-1}} = \lim_{x\to\infty} \frac{1}{-(e^{1/x}-1)^{-2}(\frac{1}{x}e^{1/x})}$

The problem is that the exponential terms never go away. The bigger problem is I believe L'Hopital's Rule is probably unnecessary and I'm missing something more basic here.

You could write the ratio differently.
$$\lim_{x\to\infty} \frac{e^{1/x}-1}{1/x}$$
That form succumbs straightforwardly to the hospital rule.