How do you show that the limit as x approaches negative infinity of e^{x}|x|^{a }= 0?

a > 0

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- Oct 24th 2012, 04:02 AMCinnamanLimit problem
How do you show that the limit as x approaches negative infinity of e

^{x}|x|^{a }= 0?

a > 0 - Oct 24th 2012, 04:22 AMemakarovRe: Limit problem
This thread on math.stackexchange.com may be helpful.

- Oct 24th 2012, 05:11 AMCinnamanRe: Limit problem
Well in a previous problem they asked me to show that a) lim

_{x->0}x^{a}ln(x) = 0 (a > 0)

That was pretty straight forward using L'hopital's rule on lim_{x->0}ln(x) / x^{-a}

But I'm having troubles with the followup question b) lim_{x->- inf}|x|^{a}e^{x} = 0

They say: by putting x = ln (t) in a), show b)

(?) - Oct 24th 2012, 06:51 AMemakarovRe: Limit problem
We have the chain rule for limits, which says that for , if

- ,
- , and either

- or
- g does not take the value d near c

then . In fact, c, d and e can also have the form or for ; in this case, the rule refers to the corresponding one-sided limit.

We need to prove where a > 0. Let b = 1 / a.

We know that where . Let . Then and for all x. By the chain rule, and . Therefore, .

Next, let . Then . By the chain rule again, and .