# Limit evaluation problem

I've got this limit and I can't figure out the right steps to find the result. (I know the result).

$\lim_{x\to-\infty}e^{x}(x+1)^{n}, n\in \Bbb N$

Any hints?

-

We apply the L'Hôpital's rule $n$ times we get:

$$\lim_{x\to-\infty}e^x(x+1)^n=\lim_{x\to-\infty}\frac{(x+1)^n}{e^{-x}}\\=-n\lim_{x\to-\infty}\frac{(x+1)^{n-1}}{e^{-x}}=\cdots=(-1)^nn!\lim_{x\to-\infty}\frac{1}{e^{-x}}=0$$

-
That's great. I did try LH but just once and I didn't think about it that way :) thanks. – Breldor Aug 16 '14 at 13:01
You're welcome. – user63181 Aug 16 '14 at 13:03
I never liked the brute force method. It would be nice to see through the limiting process. – Troy Woo Aug 16 '14 at 13:19

$$\lim_{x \to -\infty} e^x(x+1)^n=\lim_{x \to -\infty} \frac{(x+1)^n}{e^{-x}} \overset{DLH}{=} \lim_{x \to -\infty} \frac{n(x+1)^{n-1}}{-e^{-x}}\overset{DLH}{=} \\ \lim_{x \to -\infty} \frac{n(n-1)(x+1)^{n-2}}{e^{-x}}\overset{DLH}{=} \dots \overset{DLH}{=} \lim_{x \to -\infty} \frac{n!}{(-1)^ne^{-x}}=(-1)^n \lim_{x \to -\infty} e^x=0$$

-

$$\lim_{x\to-\infty}e^{x}(x+1)^{n}, n\in N$$

$$=\lim_{x\to-\infty}e^{x}x^n(1+\frac 1x)^{n}=0$$ as $$\lim_{x\to-\infty}e^x=0$$[Note that n is constant]

-