Question

How do you find `lim (sqrt(y+1)+sqrt(y-1))/y` as `y->oo` using l'Hospital's Rule?

Answer 1

`lim_(y->oo)(sqrt(y+1)+sqrt(y-1))/y=0`

Explanation:

L'Hospital's Rule is applicable if in a limit `lim_(x->a)(f(x))/(g(x))`, either as `x->a`, `f(x)` and `g(x)` both tend to `0` or to `oo`. In such cases `lim_(x->a)(f(x))/(g(x))=(f'(a))/(g'(a))`. If `(f'(a))/(g'(a))=oo/oo` or `0/0`, then we go for `(f''(a))/(g''(a))`

Here we have `y` and as `y->oo`, both numerator and denominator tend to `oo`.

Hence `lim_(y->oo)(sqrt(y+1)+sqrt(y-1))/y`

= `lim_(y->oo)(1/(2sqrt(y+1))+1/(2sqrt(y-1)))/1`

= `(1/(2xxoo)+1/(2xxoo))/1`

= `0`