How do you show that the limit as x approaches negative infinity of e^{x}|x|^{a }= 0?
a > 0
This thread on math.stackexchange.com may be helpful.
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)
(?)
We have the chain rule for limits, which says that for $\displaystyle c,d,e\in\mathbb{R}\cup\{-\infty,\infty\}$, if
- $\displaystyle \lim_{y \to d} f(y) = e$,
- $\displaystyle \lim_{x \to c} g(x) = d$, and either
- $\displaystyle f(d)=e$ or
- g does not take the value d near c
then $\displaystyle \lim_{x \to c} f(g(x)) = e$. In fact, c, d and e can also have the form $\displaystyle +a$ or $\displaystyle -a$ for $\displaystyle a\in\mathbb{R}$; in this case, the rule refers to the corresponding one-sided limit.
We need to prove $\displaystyle \lim_{x\to-\infty}e^x|x|^a = 0$ where a > 0. Let b = 1 / a.
We know that $\displaystyle \lim_{y\to+0}f(y) = 0$ where $\displaystyle f(y)=y^b\ln y$. Let $\displaystyle g(x)=e^x$. Then $\displaystyle \lim_{x\to-\infty}g(x)=+0$ and $\displaystyle e^x\ne0$ for all x. By the chain rule, $\displaystyle \lim_{x\to-\infty}f(g(x))=0$ and $\displaystyle f(g(x))=e^{bx}x$. Therefore, $\displaystyle \lim_{x\to-\infty}e^{bx}|x|=+0$.
Next, let $\displaystyle h(z)=z^a$. Then $\displaystyle \lim_{z\to+0}h(z)=0$. By the chain rule again, $\displaystyle \lim_{x\to-\infty}h(e^{bx}|x|)=0$ and $\displaystyle h(e^{bx}|x|)=e^{abx}|x|^a=e^x|x|^a$.