How to do Delta Epsilon proofs on limits that need to use L'Hopitals rule? Hi I'm an 8th grade self studying math and I was curious how would you prove a limit with the $\delta - \epsilon $ that as far as I know needs to use L'Hopitals rule. The limit that I was trying to proof was this:
$$\lim_{x\to 0^+}\frac{\frac{1}{x}}{e^{\frac{1}{x}}}=0$$
After using L'Hopitals Rule I would like to proof it with the definition but i got stuck. This is How far I got:
$\forall\epsilon>0 \exists \delta$ which is equal to ...(I don't know yet I'm trying to find it)
Suppose $0<x<\delta$
Check :
$|\frac{\frac{1}{x}}{e^{\frac{1}{x}}}|<...$(I don't know how to manipulate this equation)
So Im stuck is there a way to solve this? Do i need to do L'Hopitals rule in the absolute value? Sorry for the broken English. It isn't my first language
 A: If you want to avoid De L'Hôpital, you might do the following. First, show that $e^x\geq \frac{x^2}{2}$ for all $x\geq 0$. Indeed you have that

*

*$e^x\geq 0$ for all $x\geq 0$;


*$(e^x)' = e^x$ for all $x\geq 0$;


*as a consequence of (1) and (2) $e^x$ is increasing for $x\geq 0$ and so $e^x\geq e^0=1$ for all $x\geq 0$;


*in particular, $e^x\geq 1$ for all $x\geq 0$;


*it follows that $(e^x - x)' = e^x-1\geq 0$ for all $x\geq 0$;


*hence $e^x - x$ is increasing for all $x\geq 0$ and in particular $e^x - x \geq 1$ for all $x\geq 0$;


*now you have that $(e^x -\frac{x^2}{2})' = e^x-x\geq 1$ for all $x\geq 0$;


*so $e^x - \frac{x^2}{2}$ is increasing for $x\geq 0$ and in particular $e^x\geq \frac{x^2}{2}$ for all $x\geq 0$.
Now, writing $1/x$ instead of $x$, we have just shown that for all $x>0$ we have
$$e^{1/x}\geq \frac{1}{2x^2}\,.$$
Let's go back to your limit. Clearly for $x\geq 0$ both $1/x$ and $e^{1/x}$ are non negative and so
$$\lim_{x\to0^+}\frac{1/x}{e^{1/x}}\geq 0\,.$$
On the other hand, from what we have shown,
$$\lim_{x\to0^+}\frac{1/x}{e^{1/x}} \leq \lim_{x\to0^+}\frac{1/x}{1/(2x^2)} = \lim_{x\to0^+}\frac{2x^2}{x} = 0\,.$$
Actually we have shown something a bit stronger, as we have that for all $x > 0$
$$\left|\frac{1/x}{e^{1/x}}\right|\leq 2x\,.$$
So if you really want to use the $\epsilon$ and $\delta$ you can sat that for all $\epsilon>0$, it is enough to choose $\delta = \epsilon/2$ to ensure that if $x\in(0, \delta)$ you have that $\left|\frac{1/x}{e^{1/x}}\right|\leq \epsilon$.
