# Math Help - Limit problem

1. ## Limit problem

How do you show that the limit as x approaches negative infinity of ex|x|a = 0?

a > 0

3. ## Re: Limit problem

Well in a previous problem they asked me to show that a) limx->0 xaln(x) = 0 (a > 0)

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

But I'm having troubles with the followup question b) limx->- inf |x|aex​ = 0

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

(?)

4. ## Re: Limit problem

We have the chain rule for limits, which says that for $c,d,e\in\mathbb{R}\cup\{-\infty,\infty\}$, if

1. $\lim_{y \to d} f(y) = e$,
2. $\lim_{x \to c} g(x) = d$, and either
1. $f(d)=e$ or
2. g does not take the value d near c

then $\lim_{x \to c} f(g(x)) = e$. In fact, c, d and e can also have the form $+a$ or $-a$ for $a\in\mathbb{R}$; in this case, the rule refers to the corresponding one-sided limit.

We need to prove $\lim_{x\to-\infty}e^x|x|^a = 0$ where a > 0. Let b = 1 / a.

We know that $\lim_{y\to+0}f(y) = 0$ where $f(y)=y^b\ln y$. Let $g(x)=e^x$. Then $\lim_{x\to-\infty}g(x)=+0$ and $e^x\ne0$ for all x. By the chain rule, $\lim_{x\to-\infty}f(g(x))=0$ and $f(g(x))=e^{bx}x$. Therefore, $\lim_{x\to-\infty}e^{bx}|x|=+0$.

Next, let $h(z)=z^a$. Then $\lim_{z\to+0}h(z)=0$. By the chain rule again, $\lim_{x\to-\infty}h(e^{bx}|x|)=0$ and $h(e^{bx}|x|)=e^{abx}|x|^a=e^x|x|^a$.