# Find the limits: lim_{xrightarrow4}frac{4x-x^2}{2-sqrt{x}}

geduiwelh 2020-12-16 Answered
Find the limits: $\underset{x\to 4}{lim}\frac{4x-{x}^{2}}{2-\sqrt{x}}$
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## Expert Answer

estenutC
Answered 2020-12-17 Author has 81 answers
We have to find the limits:
$\underset{x\to 4}{lim}\frac{4x-{x}^{2}}{2-\sqrt{x}}$
It is $\frac{0}{0}$ form
Simplifying the limit by using rationalization of denominator.
Multiplying and dividing by $\left(2+\sqrt{x}\right),$
$\underset{x\to 4}{lim}\frac{4x-{x}^{2}}{2-\sqrt{x}}×\frac{\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)}$
We know the idenity,
$\left(a-b\right)\left(a+b\right)={a}^{2}-{b}^{2}$
Therefore,
$\underset{x\to 4}{lim}\frac{4x-{x}^{2}}{2-\sqrt{x}}×\frac{\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)}=\underset{x\to 4}{lim}\frac{\left(4x-{x}^{2}\right)\left(2+\sqrt{x}\right)}{\left(2{\right)}^{2}-\left(\sqrt{x}{\right)}^{2}}$

$\underset{x\to 4}{lim}x\left(2+\sqrt{x}\right)$
$=4\left(2+\sqrt{4}\right)$
$=4\left(2+2\right)$
$=4×4$
$=16$
Hence, value of limit is 16.
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Jeffrey Jordon
Answered 2022-04-01 Author has 2087 answers

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