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.

\(\lim_{x\rightarrow4}\frac{4x-x^2}{2-\sqrt{x}}\)

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.