# lim_(x rarr 4) (3 - sqrt(5 + x))/(1- sqrt(5 - x)) = ?

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Mar 9, 2018

The limit should approach -1/3, I screwed up the original answer.

#### Explanation:

${\lim}_{x \rightarrow 4} \frac{3 - \sqrt{5 + x}}{1 - \sqrt{5 - x}}$

first multiply the top and bottom by the conjugate of the numerator and the conjugate of the denominator

$\frac{3 - \sqrt{5 + x}}{1 - \sqrt{5 - x}} \cdot$

$\frac{3 + \sqrt{5 + x}}{3 + \sqrt{5 + x}} \cdot \frac{1 + \sqrt{5 - x}}{1 + \sqrt{5 - x}}$

$= \frac{4 - x}{- \left(4 - x\right)} \cdot \frac{1 + \sqrt{5 - x}}{3 + \sqrt{5 + x}}$

$= - \frac{1 + \sqrt{5 - x}}{3 + \sqrt{5 + x}}$

plug in the limit value to get your answer:

$= - \frac{1 + \sqrt{5 - 4}}{3 + \sqrt{5 + 4}}$
$= - \frac{1 + 1}{3 + 3} = - \frac{2}{6} = - \frac{1}{3}$

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Jim H Share
Mar 8, 2018

The limit is $- \frac{1}{3}$

#### Explanation:

It is similar to both

${\lim}_{x \rightarrow 4} \frac{3 - \sqrt{5 + x}}{x - 4}$ and to ${\lim}_{x \rightarrow 4} \frac{x - 4}{1 - \sqrt{5 - x}}$

but both in one expression.

So multiply $\frac{3 - \sqrt{5 + x}}{1 - \sqrt{5 - x}}$ by

$\frac{\left(3 + \sqrt{5 + x}\right)}{\left(3 + \sqrt{5 + x}\right)} \cdot \frac{\left(1 + \sqrt{5 - x}\right)}{\left(1 + \sqrt{5 - x}\right)}$ to get:

lim_(xrarr4)((9-(5+x))(1+sqrt(5-x)))/((3+sqrt(5+x))(1-(5-x))

 = lim_(xrarr4)((4-x)(1+sqrt(5-x)))/((3+sqrt(5+x))(-(4-x))

$= {\lim}_{x \rightarrow 4} \frac{- \left(1 + \sqrt{5 - x}\right)}{3 + \sqrt{5 + x}}$

$= \frac{- \left(1 + \sqrt{1}\right)}{3 + \sqrt{9}} = - \frac{2}{6} = - \frac{1}{3}$

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