Question

How do you find `lim sqrt(x+2)-sqrtx` as `x->oo`?

Answer 1

`lim_(xrarroo)(sqrt(x+2)-sqrtx)` has indeterminate initial form `oo-oo`. See below.

Explanation:

`(sqrt(x+2)-sqrtx) = ((sqrt(x+2)-sqrtx))/1 ((sqrt(x+2)+sqrtx))/((sqrt(x+2)+sqrtx))`

`= (x+2-x)/(sqrt(x+2)+sqrtx)`

`= 2/(sqrt(x+2)+sqrtx)`

As `xrarroo`, the denominator `sqrt(x+2)+sqrtx rarr oo`

Therefore,

`lim_(xrarroo) (sqrt(x+2)-sqrtx) =lim_(xrarroo) 2/(sqrt(x+2)+sqrtx)`

`= 0`