Question

How do you find the limit `x^2/(e^x-x-1)` as `x->0`?

Answer 1

`2`

Explanation:

Remembering that

`e^x=1+x+x^2/(2!)+x^3/(3!)+cdots`

`x^2/(e^x-x-1)=x^2/(x^2/(2!)+x^3/(3!)+cdots)=1/(1/(2!)+x/(3!)+cdots)` so

`lim_(x->0)x^2/(e^x-x-1)=lim_(x->0)1/(1/(2!)+x/(3!)+cdots)=2`