Evaluate $\lim_{x\to\infty} \frac{xe^x}{e^{x^2}}$.

Question

Evaluate $$\lim_{x\to\infty} \frac{xe^x}{e^{x^2}}.$$ I tried the L’Hopital’s rule but got nothing. Any help would be great!

Try L'hopital's on $$\dfrac{x}{e^{x^2-x}}$$ – AgentS Aug 03 '19 at 07:48 — AgentS, Aug 03 '19 at 07:48
You could have posted this as an answer @rsadhvika – 19aksh Aug 03 '19 at 07:55 — 19aksh, Aug 03 '19 at 07:55

score 3 · Answer 1 · answered Aug 03 '19 at 07:56

3

Let $x>2$:

Note: $e^{x^2} > e^{2x}$, since $x^2>2x$.

$0 < \dfrac{xe^x}{e^{x^2}} < \dfrac{x}{e^x} $;

Take the limit.

answered Aug 03 '19 at 07:56

Peter Szilas

I believe your method is that of the application of Squeeze Theorem, right? – Nzewi Ernest Kenechukwu Aug 03 '19 at 11:20
Nzewi Ernest Kenechukwu.Yes. Limit of LHS =0, limit of RHS = 0 , hence the limit in "the middle" =0; OK? – Peter Szilas Aug 03 '19 at 12:31

score 2 · Answer 2 · answered Aug 03 '19 at 07:51

2

Hint. Note that $$\frac{xe^x}{e^{x^2}}=\exp(-x^2+x+\ln(x))=\exp\left(-x^2\left(1-\frac{1}{x}-\frac{\ln(x)}{x^2}\right)\right).$$ Then recall Proving $\lim_{n\to \infty} \frac{(\ln n)^a}{n^b} = 0$ for all $a,b>0$

answered Aug 03 '19 at 07:51

Robert Z

score 1 · Accepted Answer · answered Aug 03 '19 at 07:49

1

${{xe^x}\over{e^{x^2}}}$

$={x\over{e^{x^2-x}}}$

Hospital ${1\over{(2x-1)e^{x^2-x}}}$ the limit is zero

answered Aug 03 '19 at 07:49

Tsemo Aristide

3 Answers3