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

Question
Limits and continuity
asked 2020-12-16
Find the limits: \(\lim_{x\rightarrow4}\frac{4x-x^2}{2-\sqrt{x}}\)

Answers (1)

2020-12-17
We have to find the limits:
\(\lim_{x\rightarrow4}\frac{4x-x^2}{2-\sqrt{x}}\)
It is \(\frac{0}{0}\) form
Simplifying the limit by using rationalization of denominator.
Multiplying and dividing by \((2+\sqrt{x}),\)
\(\lim_{x\rightarrow4}\frac{4x-x^2}{2-\sqrt{x}}\times\frac{(2+\sqrt{x})}{(2+\sqrt{x})}\)
We know the idenity,
\((a-b)(a+b)=a^2-b^2\)
Therefore,
\(\lim_{x\rightarrow4}\frac{4x-x^2}{2-\sqrt{x}}\times\frac{(2+\sqrt{x})}{(2+\sqrt{x})}=\lim_{x\rightarrow4}\frac{(4x-x^2)(2+\sqrt{x})}{(2)^2-(\sqrt{x})^2}\)
\(\lim_{x\rightarrow4}\frac{x(4-x)(2+\sqrt{x})}{4-x}\ (\text{taken x common from }4x-x^2)\)
\(\lim_{x\rightarrow4}x(2+\sqrt{x})\)
\(=4(2+\sqrt{4})\)
\(=4(2+2)\)
\(=4\times4\)
\(=16\)
Hence, value of limit is 16.
0

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