Is the Limit of x^n/e^x Equal to 0 for Negative Integers?

[phymatter]

is lim_x->infinity xⁿ/e^x =0 for n = -ve integers , if yes then why?
also is (->infinity)⁰ =1 ?

 

[Lodve]

You know if you get inifinity over infinity you must "change" your fractionexpression by using L'hopitas rule. It says that if you either get infinity over ifinity or zero over zero you must derive nominator og denominator so that you get a fractionexpression which is capable of giving you an answer by substituting the variable x.

 

[mathman]

is lim_x->infinity xⁿ/e^x =0 for n = -ve integers , if yes then why?
also is (->infinity)⁰ =1 ?

I don't know what you mean by n = -ve integers. However using L'Hopital's rule for any fixed n, the result will always be 0. e^x will persist as you take derivatives, while xⁿ will be 0 after n derivatives.

 

[phymatter]

I don't know what you mean by n = -ve integers.

by n = -ve integers i mean that n={-1,-2,-3,-4,...}

 

[HallsofIvy]

is lim_x->infinity xⁿ/e^x =0 for n = -ve integers , if yes then why?
also is (->infinity)⁰ =1 ?

Surely it is not that difficult to write "negative"!

In any case, it should be obvious that x^n/e^x, for n negative, is the same as 1/(x^ne^x) with n positive. And since both x^n and e^x grow without bound as x goes to infinity, the fraction goes to 0.

Again (->infinity) makes no sense. If you mean $\lim_{x\to\infty}f(x)$ it still makes no sense because that limit might be 0, in which case $0^0$ is not defined, or that limit might be a non-zero number, in which case the limit is 1.