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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?

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    make a substitution e.g. $$t=\frac{1}{x}$$ – 2017-02-23
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    You need to know $$\lim_{x\to+\infty} \frac{e^x}{x^\alpha}$$ this is often done with l'Hopital, though. – 2017-02-23
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    @G.Sassatelli if $\lim_{x\to\infty} \frac{x^\alpha}{e^x} = 0$, is $\lim_{x\to\infty} \frac{e^x}{x^\alpha} = \infty$ ? – 2017-02-23
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    @MarkRead Indeed, with $+\infty$ on both sides, though. – 2017-02-23
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    @G.Sassatelli Probably a silly question, but what do you mean exactly by both sides? – 2017-02-23

3 Answers 3

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Without using L'Hopital but although it looks same,

$x=\dfrac{1}{t}\\$ , $\displaystyle L = \lim_{t\to\infty}\dfrac{e^t}{t^2} = \lim_{t\to\infty}\dfrac{1+t+\dfrac{t^2}{2!}+\cdots}{t^2} = \dfrac{1}{2}+ \lim_{t\to\infty}\left(\dfrac{t}{3!}+\dfrac{t^2}{4!}+\cdots\right)=\infty$

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$$\lim_{x\to 0^+} x^2 e^{1/x} \stackrel{x\mapsto 1/y}{=} \lim_{y\to +\infty}\frac{e^y}{y^2}=\lim_{y\to +\infty}\frac{(e^{y/3})^3}{y^2}\geq \lim_{y\to +\infty}\frac{\left(1+\frac{y}{3}\right)^3}{y^2}=+\infty. $$

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You can expand the exponential term in series by remembering that \begin{equation} 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!} \end{equation} You can then multiply by $x²$ and write the apply the limit. The limit goes to $+\infty$.

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    Limits don't diverge; they either exist or don't. – 2017-02-23