Question

How do you find the limit of `(2x-8)/(sqrt(x) -2)` as x approaches 4?

Answer 1

`8`

Explanation:

As you can see, you will find an indeterminate form of `0/0` if you try to plug in `4`. That is a good thing because you can directly use L'Hospital's Rule, which says

`if lim_(x ->a) (f(x))/(g(x))=0/0 or oo/oo`

all you have to do is to find the derivative of the numerator and the denominator separately then plug in the value of `x`.

`=>lim_(x->a)(f'(x))/(g'(x)`

`f(x)= lim_(x->4) (2x-8)/(sqrtx-2)=0/0`

`f(x)= lim_(x->4)(2x-8)/(x^(1/2)-2)`

`f'(x)=lim_(x->4) (2)/(1/2x^(-1/2)) = lim_(x->4)(2)/(1/(2sqrtx))=(2)/(1/4)=8`

Hope this helps :)

Answer 2

`lim_(x->4)(2x-8)/(sqrt(x)-2)=8`

Explanation:

As an addition to the other answer, this problem can be solved by applying algebraic manipulation to the expression.

`lim_(x->4)(2x-8)/(sqrt(x)-2) = lim_(x->4)2*(x-4)/(sqrt(x)-2)`

`=lim_(x->4)2*((x-4)(sqrt(x)+2))/((sqrt(x)-2)(sqrt(x)+2))`

`=lim_(x->4)2*((x-4)(sqrt(x)+2))/(x-4)`

`=lim_(x->4)2(sqrt(x)+2)`

`=2(sqrt(4)+2)`

`=2(2+2)`

`=8`