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I have a problem with this limit, i have no idea how to compute it. Can you explain the method and the steps used? Thanks $$\lim _{x\to 0+}\left(\frac{\frac{e^{-\frac{1}{x}}}{x^3} +e^{-\frac{1}{\sqrt{x}}}}{e^{-\frac{1}{x}}lnx}\right)$$

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closed as off-topic by user296602, user228113, colormegone, Shailesh, Milo Brandt Jan 9 '16 at 1:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Community, colormegone, Shailesh, Milo Brandt
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Welcome to MSE! This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ – user296602 Jan 8 '16 at 22:20
  • $\begingroup$ You should include your thoughts on this problem. Which part of the limit confuses you? Have you tried breaking this down into smaller parts and seeing which part it is that gives you the most trouble? $\endgroup$ – Brandon Thomas Van Over Jan 8 '16 at 22:33
  • $\begingroup$ the searched Limit doesn't exist (is $-\infty$) $\endgroup$ – Dr. Sonnhard Graubner Jan 8 '16 at 23:36
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It is convenient to make the change of variables $x=1/t$ to get $$ \lim_{t\to +\infty}\frac{t^3 e^{-t}+e^{-\sqrt{t}}}{-e^{-t}\ln t}= \lim_{t\to +\infty}\left[-\frac{t^3}{\ln t}-\frac{e^{t-\sqrt{t}}}{\ln t}\right]\ . $$ The term $t^3/\ln t\to +\infty$ as any power goes to infinity faster than $\ln$, and the second term $\frac{e^{t-\sqrt{t}}}{\ln t}$ goes to $+\infty$ as $t$ wins w.r.t. $\sqrt{t}$, and the exponential goes to infinity faster than $\ln$. Therefore the requested limit exists and is equal to $-\infty$.

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