Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 7 down vote accepted

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

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

share|cite|improve this answer
    
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

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

share|cite|improve this answer

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


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

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.