If f(4) = 4, f'(4) = 1, then $\lim\limits_{x \rightarrow 4} \frac{2-\sqrt{f(x)}}{2-\sqrt{x}}$ is equal to
Answer & explanation
Correct answer: option 2
We have,
$\lim\limits_{x \rightarrow 4} \frac{2-\sqrt{f(x)}}{2-\sqrt{x}}$
$=\lim\limits_{x \rightarrow 4} \frac{4-f(x)}{4-x} \times \frac{2+\sqrt{x}}{2+\sqrt{f(x)}}$
$=\lim\limits_{x \rightarrow 4} \frac{f(x)-4}{x-4} \times \lim\limits_{x \rightarrow 4} \frac{2+\sqrt{x}}{2+\sqrt{f(x)}}$
$=f'(4) \times \frac{2+2}{2+\sqrt{f(4)}}$ $\left[\begin{array}{c}∵ f(x) \text { is differentiable at } x=4 \\ ∴ \text { It is continuous at } x = 4 . \\ \text { Hence, } \lim\limits_{x \rightarrow 4} f(x)=f(4)\end{array}\right]$
$=1 \times \frac{4}{2+2}=1$