If f(16)=16 and f'(16) = 5, then \(\displaystyle \lim _{x \rightarrow 16} \frac{\sqrt{f(x)}-4}{\sqrt{x}-4}=\text { ? }\)

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Question

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AAI ATC Junior Executive 21 Feb 2023 Shift 2 Official Paper

Answer 1

Given:

If f(16)=16 and f'(16) = 5 then

\(\displaystyle \lim _{x \rightarrow 16} \frac{\sqrt{f(x)}-4}{\sqrt{x}-4}=\text { ? } \)

Concept:

Use concept of L'Hopital rule.

Calculation:

\(\displaystyle \lim _{x \rightarrow 16} \frac{\sqrt{f(x)}-4}{\sqrt{x}-4}\)

This is \(\rm \frac{0}{0}\) form then apply L'Hopital Rule

\(\displaystyle =\lim _{x \rightarrow 16} \frac{\frac{1}{2\sqrt{f(x)}}\cdot f'(x)}{\frac{1}{2\sqrt{x}}}\)

\(\displaystyle =\lim _{x \rightarrow 16} \frac{\sqrt{x}\cdot f'(x)}{\sqrt{f(x)}}\)

\(\rm =\frac{\sqrt{16}\cdot5}{\sqrt{16}}\) (as f(16) = 16 and f'(16) = 5)

= 5

Hence the option (1) is correct.

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