What is the limit of x to infinity of x^2-3/e^x?

1 Answer
Jun 8, 2018

L=lim_(x->oo)(x^2-3)/e^x=0

Explanation:

We want to evaluate the limit

L=lim_(x->oo)(x^2-3)/e^x

Which is an indeterminate form (oo)/(oo)

So we can apply L'Hôpital's rule

color(blue)(lim_(x->c)f(x)/g(x)=lim_(x->c)(f'(x))/(g'(x))

Thus

L=lim_(x->oo)(2x)/e^x

An indeterminate form (oo)/(oo), so apply LHR again

L=lim_(x->oo)(2)/e^x=0