Question

What is the limit of `(sqrt(x^2+4x+1)-x)` as x goes to infinity?

Answer 1

`lim_(xrarroo)(sqrt(x^2+4x+1)-x) = 2`

Explanation:

`lim_(xrarroo)(sqrt(x^2+4x+1)-x)` has form `oo-oo` which is indeterminate.

`(sqrt(x^2+4x+1)-x) = ((sqrt(x^2+4x+1)-x))/1 ((sqrt(x^2+4x+1)+x))/((sqrt(x^2+4x+1)+x))`

`= (x^2+4x+1-x^2)/(sqrt(x^2+4x+1)+x)`

`= (4x+1)/(sqrt(x^2(1+4/x+1/x^2))+x)` for `x != 0`

`= (x(4+1/x))/(x(sqrt(1+4/x+1/x^2)+1))` for `x > 0`

`= (4+1/x)/(sqrt(1+4/x+1/x^2)+1)` for `x > 0`

Taking the limit as `xrarroo`, we get `4/(sqrt1+1) = 4/2=2`