# Solve $\lim_{x \rightarrow 0^+}(e^{\frac{1}{x}}x^2)$ without using L'Hopital's rule

I tried:

$\lim_{x \rightarrow 0^+}(e^{\frac{1}{x}}x^2) = x^2 \cdot \frac{1}{e^{-\frac{1}{x}}} = \frac{x^2}{e^{-\frac{1}{x}}} = ???$

I thought maybe I could use $y = - \frac{1}{x}$, but I don't know what to do next.

I know the limit just by looking a the function: $\lim_{x \rightarrow 0^+} e^{\frac{1}{x}} = \infty$ and $\lim_{x \rightarrow 0^+} x^2 \rightarrow$ values close to 0 but greater than zero. And so the answer is $\infty$ but this looks incomplete. How do I solve this analitically?

• make a substitution e.g. $$t=\frac{1}{x}$$ – Dr. Sonnhard Graubner Feb 23 '17 at 19:39
• You need to know $$\lim_{x\to+\infty} \frac{e^x}{x^\alpha}$$ this is often done with l'Hopital, though. – user228113 Feb 23 '17 at 19:42
• @G.Sassatelli if $\lim_{x\to\infty} \frac{x^\alpha}{e^x} = 0$, is $\lim_{x\to\infty} \frac{e^x}{x^\alpha} = \infty$ ? – Mark Read Feb 23 '17 at 19:46
• @MarkRead Indeed, with $+\infty$ on both sides, though. – user228113 Feb 23 '17 at 19:48
• @G.Sassatelli Probably a silly question, but what do you mean exactly by both sides? – Mark Read Feb 23 '17 at 19:49

$x=\dfrac{1}{t}\\$ , $\displaystyle L = \lim_{t\to\infty}\dfrac{e^t}{t^2} = \lim_{t\to\infty}\dfrac{1+t+\dfrac{t^2}{2!}+\cdots}{t^2} = \dfrac{1}{2}+ \lim_{t\to\infty}\left(\dfrac{t}{3!}+\dfrac{t^2}{4!}+\cdots\right)=\infty$
$$\lim_{x\to 0^+} x^2 e^{1/x} \stackrel{x\mapsto 1/y}{=} \lim_{y\to +\infty}\frac{e^y}{y^2}=\lim_{y\to +\infty}\frac{(e^{y/3})^3}{y^2}\geq \lim_{y\to +\infty}\frac{\left(1+\frac{y}{3}\right)^3}{y^2}=+\infty.$$
You can expand the exponential term in series by remembering that $$e^\frac{1}{x}= \sum_{k=0}^{+\infty} \frac{x^{-k}}{k!}=1+\frac{1}{x}+\frac{1}{2 x^2}+\sum_{k=3}^{+\infty} \frac{x^{-k}}{k!}$$ You can then multiply by $x²$ and write the apply the limit. The limit goes to $+\infty$.