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I want to evaluate

$$\lim_{x\rightarrow0^+}\frac{\log{x}}{e^{1/x}}$$

I know that for $x\rightarrow0^+$, $\log{x}\rightarrow-\infty$ and $e^{1/x}\rightarrow+\infty$.

This leads to an indeterminate form $\left[\frac{\infty}{\infty}\right]$, so I'm not sure what to do in these situations. Perhaps change the variable, but not sure what's the logic behind chaning the variable of the limit.

Any hints?

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    $\begingroup$ You can change variables like this $t=1/x \to \infty$, and use that exponential grows much faster than logarithm, or just use l'Hospital's rule. Limit is $0$ $\endgroup$ – Jakobian Aug 11 '18 at 13:21
  • $\begingroup$ Thanks @Rumpelstiltskin that worked! $\endgroup$ – Cesare Aug 11 '18 at 13:23
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Let $x=\dfrac1u$ then $$\lim_{x\rightarrow0^+}\frac{\log{x}}{e^{1/x}}=\lim_{u\to+\infty}\dfrac{-\ln u}{e^u}=\lim_{u\to+\infty}\dfrac{-\dfrac1u}{e^u}=\lim_{u\to+\infty}\dfrac{-1}{ue^u}=0$$

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$$\lim_{x\rightarrow0^+}\left(\dfrac{\log x}{e^\frac1x}\right)=\lim_{x\rightarrow0^+}\left(\dfrac{\ln x}{\dfrac{1}{e^{-\frac1x}}}\right)$$Now apply L'Hopitals rule and we get$$\lim_{x\rightarrow0^+}\left(\dfrac{-x}{e^\frac1x}\right)=-\lim_{x\rightarrow0^+}\left(\dfrac{x}{e^{\frac1x}}\right)=0$$

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