# Limit as x-> inf

Tags:
1. Feb 16, 2017

1. The problem statement, all variables and given/known data
Find $\lim_{x\to\infty} x(e^{1/x}-1)$

2. Relevant equations
$\lim_{x\to\infty} \frac{f(x)}{g(x)} = \lim_{x\to\infty} \frac{f'(x)}{g'(x)}$

3. 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.

2. Feb 16, 2017

### BvU

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

3. Feb 16, 2017

### PeroK

Another idea is to use power series.

4. Feb 17, 2017

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.

5. Feb 17, 2017

### vela

Staff Emeritus
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.