Question

How do you find the limit `lim_(xto0)(sqrt(9+x)-3)/x`?

Answer 1

Begin by evaluating the expression at `x=0` `(sqrt(9+0)-3)/(0) = 0/0`.
The indeterminate form indicates that one should use L'Hôpital's rule

Explanation:

Apply L'Hôpital's rule by differentiating both the numerator and the denominator with respect to x:

`lim_(xto0)((d(sqrt(9+x)-3))/dx)/((d(x))/dx)`

The results of the differentiation:

`lim_(xto0)(1/(2(sqrt(9+x))))/1`

Simplify:

`lim_(xto0)1/(2(sqrt(9+x)))`

Evaluate at `x=0`

`1/(2(sqrt(9+0))) = 1/6`

The rule says that the original expression goes to the same limit:

`lim_(xto0)(sqrt(9+x)-3)/x = 1/6`

Answer 2

`1/6`

Explanation:

We know that for continuous `f(x)`

`lim_(h->0)(f(x+h)-f(x))/h = f'(x)`

now calling `f(x)=sqrt(x)` we have

`lim_(x->0)(sqrt(9+x)-sqrt(9))/x = 1/2 (1/sqrt9) = 1/6`