Question

How do you find `lim sqrtx+sqrt(x+1)` as `x->oo`?

Answer 1

`lim_(x->oo) sqrtx+ sqrt(x+1) = +oo`

Explanation:

We have:

`lim_(x->oo) sqrtx = +oo`

and

`lim_(x->oo) sqrt(x+1) = +oo`

So:

`lim_(x->oo) sqrtx+ sqrt(x+1) = +oo`