# Unable to calculate limit without getting an indeterminate form

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

• How many times have you differentiated?
– LHF
Mar 15, 2020 at 15:01
• @Hadi Your edit replaced $e^{\color{red}{-}1/x}$ with $e^{1/x}$. Destructive ;) Mar 17, 2020 at 7:09
• @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.
Mar 17, 2020 at 17:02
• @Hadi: I've brought the "$-$" back. Answer (new or edited) is the right place. Mar 17, 2020 at 17:36
Mar 17, 2020 at 17:48

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 $$k\in\mathbb{N}\cup\{0\}$$.

• Use \left(\right) Mar 15, 2020 at 17:42
• @ms._VerkhovtsevaKatya thank you!
Mar 15, 2020 at 17:47
• You're welcome! Mar 15, 2020 at 19:26
• Thanks a lot, but are you sure that this is right? The solution should be zero! Mar 16, 2020 at 13:55
• @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?
$$\lim_{y\to+\infty}(k\ln(y)-y)$$
$$\lim_{y\to+\infty}y(k\frac{\ln(y)}{y}-1)=-\infty$$