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

Oct 8, 2016

$8$

#### Explanation:

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

$\mathmr{if} {\lim}_{x \to a} \frac{f \left(x\right)}{g \left(x\right)} = \frac{0}{0} \mathmr{and} \frac{\infty}{\infty}$

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)

$f \left(x\right) = {\lim}_{x \to 4} \frac{2 x - 8}{\sqrt{x} - 2} = \frac{0}{0}$

$f \left(x\right) = {\lim}_{x \to 4} \frac{2 x - 8}{{x}^{\frac{1}{2}} - 2}$

$f ' \left(x\right) = {\lim}_{x \to 4} \frac{2}{\frac{1}{2} {x}^{- \frac{1}{2}}} = {\lim}_{x \to 4} \frac{2}{\frac{1}{2 \sqrt{x}}} = \frac{2}{\frac{1}{4}} = 8$

Hope this helps :)

Oct 8, 2016

${\lim}_{x \to 4} \frac{2 x - 8}{\sqrt{x} - 2} = 8$

#### Explanation:

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

${\lim}_{x \to 4} \frac{2 x - 8}{\sqrt{x} - 2} = {\lim}_{x \to 4} 2 \cdot \frac{x - 4}{\sqrt{x} - 2}$

$= {\lim}_{x \to 4} 2 \cdot \frac{\left(x - 4\right) \left(\sqrt{x} + 2\right)}{\left(\sqrt{x} - 2\right) \left(\sqrt{x} + 2\right)}$

$= {\lim}_{x \to 4} 2 \cdot \frac{\left(x - 4\right) \left(\sqrt{x} + 2\right)}{x - 4}$

$= {\lim}_{x \to 4} 2 \left(\sqrt{x} + 2\right)$

$= 2 \left(\sqrt{4} + 2\right)$

$= 2 \left(2 + 2\right)$

$= 8$