How to do Delta Epsilon proofs on limits that need to use L'Hopitals rule?

Question

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

Answer 1

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

1. $e^x\geq 0$ for all $x\geq 0$;

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

3. 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 $x\geq 0$;

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

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

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

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

8. 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 $\left|\frac{1/x}{e^{1/x}}\right|\leq \epsilon$.