# Can't seem to figure out this limit

## Homework Statement

I'm trying to do this limit based on a previous thread ( https://www.physicsforums.com/threads/proving-n-x-n-e-x-integrated-from-0-to-infinity.641947/#_=_ )

I got up to the last part of thread where I need to find the limit of:
limit as x approaches infinity of: (-x^(k+1))/e^x

## The Attempt at a Solution

I know that this limit somehow must equal to zero in order to get the right answer, but I did l'Hopital's rule 4 times and it just keeps on going to infinity.

I attached the working out of the whole problem

Really appreciate it if someone could help

#### Attachments

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## Answers and Replies

LCKurtz
Science Advisor
Homework Helper
Gold Member
If you keep using L'Hospitals rule with a polynomial in the numerator and an exponential in the denominator, the numerator's degree will eventually become 0 while the exponential remains in the denominator.

If you keep using L'Hospitals rule with a polynomial in the numerator and an exponential in the denominator, the numerator's degree will eventually become 0 while the exponential remains in the denominator.

I cant seem to get it to work here because the exponent of the polynomail has a degree k. If the degree of the polynomail is a variable constant im not sure how i can get it to zero

I cant seem to get it to work here because the exponent of the polynomail has a degree k. If the degree of the polynomail is a variable constant im not sure how i can get it to zero

What's the derivative of any constant?

pasmith
Homework Helper

## Homework Statement

I'm trying to do this limit based on a previous thread ( https://www.physicsforums.com/threads/proving-n-x-n-e-x-integrated-from-0-to-infinity.641947/#_=_ )

I got up to the last part of thread where I need to find the limit of:
limit as x approaches infinity of: (-x^(k+1))/e^x

## The Attempt at a Solution

I know that this limit somehow must equal to zero in order to get the right answer, but I did l'Hopital's rule 4 times and it just keeps on going to infinity.

Keep going. You must apply l'Hopital $k + 1$ times in total before you get a constant in the numerator.

Alternatively, as every term in the series $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$ is strictly positive when $x > 0$, we have $e^x > \frac{x^{k+2}}{(k+2)!}$ and hence $$0 < \frac{x^{k+1}}{e^x} < \frac{(k+2)!x^{k+1}}{x^{k+2}} = \frac{(k+2)!}{x}.$$ Now use the squeeze theorem.

• StoneTemplePython
StoneTemplePython
Science Advisor
Gold Member
Keep going. You must apply l'Hopital $k + 1$ times in total before you get a constant in the numerator.

Alternatively, as every term in the series $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$ is strictly positive when $x > 0$, we have $e^x > \frac{x^{k+2}}{(k+2)!}$ and hence $$0 < \frac{x^{k+1}}{e^x} < \frac{(k+2)!x^{k+1}}{x^{k+2}} = \frac{(k+2)!}{x}.$$ Now use the squeeze theorem.

This is how I'd do it, as I tend to think l'Hopital is a rather unintuitive power-tool that should be used as a last resort.

At a minimum, I'd change it to

##0 \leq \frac{x^{k+1}}{e^x} \leq ...##

though, otherwise that strict inequality would seem to cause problems, as in the limit we have ## 0 \lt 0##