Question

How do you find `lim (sqrt(x^2+1)-1)/(sqrt(x+1)-1)` as `x->0` using l'Hospital's Rule or otherwise?

Answer 1

`0`

Explanation:

`(1+x^n)^m=1+m x^n+(m(m-1))/(2!)x^(2n)+cdots` so

`(1+x^n)^m = 1+m x^ng(x^n)` where

`g(x^n) = 1+(m-1)/(2!)x^n+((m-1)(m-2))/(3!)x^(2n)+ cdots`

In our case

`(sqrt(x^2+1)-1)/(sqrt(x+1)-1)equiv(1+1/2x^2 g(x^2)-1)/(1+1/2x g(x)-1) = (1/2x^2 g(x^2))/(1/2x g(x)) =xg(x^2)/(g(x))` so

`lim_(x->0) (sqrt(x^2+1)-1)/(sqrt(x+1)-1)=lim_(x->0) x g(x^2)/(g(x))` but `g(0)=1` so
`lim_(x->0) (sqrt(x^2+1)-1)/(sqrt(x+1)-1)=0`