Question

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

Answer 1

`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`