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

Question
Limits and continuity
Find the limits: $$\lim_{x\rightarrow4}\frac{4x-x^2}{2-\sqrt{x}}$$

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.

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