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I initially need to solve this: $$\lim_{x\to 0^+} \frac{e^{-1/x}}{x^k},\; \text{ where }k\in\mathbb{N}\cup\{0\}.$$

When I substitute $y=\dfrac1x$ then I get: $\lim\limits_{y\to\infty} \dfrac{y^k}{e^y} $

I'm unable to calculate this limit. Whatever I do, I get an indeterminate limit, even with L'Hospital's rule

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    $\begingroup$ How many times have you differentiated? $\endgroup$
    – LHF
    Mar 15, 2020 at 15:01
  • $\begingroup$ @Hadi Your edit replaced $e^{\color{red}{-}1/x}$ with $e^{1/x}$. Destructive ;) $\endgroup$
    – metamorphy
    Mar 17, 2020 at 7:09
  • $\begingroup$ @metamorphy ohhhh I'm sorry about that. The answer should be 0 then. I'll delete my previous answer and add a new one if you'd like, or I could explain it here in the comments. $\endgroup$
    – Hadi
    Mar 17, 2020 at 17:02
  • $\begingroup$ @Hadi: I've brought the "$-$" back. Answer (new or edited) is the right place. $\endgroup$
    – metamorphy
    Mar 17, 2020 at 17:36
  • $\begingroup$ @metamorphy I've edited my answer. Sorry about that! $\endgroup$
    – Hadi
    Mar 17, 2020 at 17:48

2 Answers 2

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For $k\neq0$ we have $$\displaystyle\lim_{x\to0^+}\frac{e^{-1/x}}{x^k}$$ $$=\displaystyle\lim_{x\to0^+}\frac{1}{e^{1/x}x^k}$$ $$=\left(\displaystyle\lim_{x\to0^+}\frac{1}{e^{1/x}}\right)\left(\displaystyle\lim_{x\to0^+}\frac{1}{x^k}\right)$$ $$=\left(\frac{1}{e^{\infty}}\right)\left({\frac{1}{\infty^k}}\right)$$ $$=0\times0$$ $$=0$$

In the case where $k=0$ we'd get the same answer since the expression becomes $$\displaystyle\lim_{x\to0^+}{e^{-1/x}}=e^{-\infty}=0.$$

Therefore $\displaystyle\lim_{x\to0^+}\frac{e^{-1/x}}{x^k}=0$ for all $k\in\mathbb{N}\cup\{0\}$.

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  • $\begingroup$ Use \left(\right) $\endgroup$
    – PinkyWay
    Mar 15, 2020 at 17:42
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    $\begingroup$ @ms._VerkhovtsevaKatya thank you! $\endgroup$
    – Hadi
    Mar 15, 2020 at 17:47
  • $\begingroup$ You're welcome! $\endgroup$
    – PinkyWay
    Mar 15, 2020 at 19:26
  • $\begingroup$ Thanks a lot, but are you sure that this is right? The solution should be zero! $\endgroup$ Mar 16, 2020 at 13:55
  • $\begingroup$ @smalllearner I think so. If you graph the function you can see that it tends to infinity when it approaches 0 from the right (unless I need to touch up on my left-right sided limits). Is the 0 answer part of your instructor’s/textbook’s answer? $\endgroup$
    – Hadi
    Mar 16, 2020 at 14:06
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Let compute the limit of its logarithm.

$$\lim_{y\to+\infty}(k\ln(y)-y)$$

which becomes

$$\lim_{y\to+\infty}y(k\frac{\ln(y)}{y}-1)=-\infty$$

so, your limit is zero.

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