# Calculate this limit with L’Hôpital’s Rule

calculuslimits

I want to calculate the limit $$\lim_{x\rightarrow0}x^2e^{1/x}$$, but I have no idea how this limit is supposed to calculated.

EDIT: After writing it as $$\lim_{x\rightarrow0}\frac{e^{\frac{1}{x}}}{x^{-2}}$$, use L'Hôpital's Rule twice:

$$\lim_{x\rightarrow0}\frac{(e^{\frac{1}{x}})'}{(x^{-2})'} = \lim_{x\rightarrow0}\frac{xe^{\frac{1}{x}}}{2}$$

$$\lim_{x\rightarrow0}\frac{(xe^{\frac{1}{x}})'}{(2)'} = \lim_{x\rightarrow0}\frac{e^{\frac{1}{x}}-\frac{e^{\frac{1}{x}}}{x}}{0}$$

So, as my math level (low) tell me a thing $$e^{\frac{1}{x}} > \frac{e^{\frac{1}{x}}}{x}$$ and $$e^{\frac{1}{x}} – \frac{e^{\frac{1}{x}}}{x} > 0$$

Does this mean answer is $$\inf$$? Or I just failed?

The limit, as it has been proposed, does not exist.

However both side limits exist.

Indeed, if we make the change of variable $$y = 1/x$$ and approach $$0$$ from the right, it results that \begin{align*} \lim_{x\to 0;x\in(0,\infty)}x^{2}e^{1/x} = \lim_{y\to+\infty}\frac{e^{y}}{y^{2}} = +\infty \end{align*}

On the other hand, if we make the same change of variable, but approach $$0$$ from the left, we get \begin{align*} \lim_{x\to 0;x\in(-\infty,0)}x^{2}e^{1/x} = \lim_{y\to-\infty}\frac{e^{y}}{y^{2}} = 0 \end{align*}

and we are done.

Hopefully this helps !