IIT JEE 1983 Maths - MCQ Question 16

$\lim_{x\to 1} \frac{\sqrt{24}-\sqrt{25-x^2}}{x-1}=\lim_{x\to 1} \frac{\sqrt{24}-\sqrt{25-x^2}}{x-1}\cdot \frac{\sqrt{24}+\sqrt{25-x^2}}{\sqrt{24}+\sqrt{25-x^2}}=\lim_{x\to 1} \frac{x^2-1}{x-1}\cdot \frac{1}{\sqrt{24}+\sqrt{25-x^2}}=\lim_{x\to 1} \frac{x+1}{\sqrt{24}+\sqrt{25-x^2}}=\frac{1}{\sqrt{24}}$

$\begin{aligned} L & = \lim_{x \to 1} \frac {G(x)-G(1)}{x-1} & \small \color{#3D99F6} \text{Given that } G(x) = - \sqrt{25-x^2} \\ & = \lim_{x \to 1} \frac {- \sqrt{25-x^2} + \sqrt{25-1}}{x-1} & \small \color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.} \\ & = \lim_{x \to 1} \frac {-\frac {-2x}{2\sqrt{25-x^2}}}1 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \frac 1{\sqrt{24}} \end{aligned}$

Therefore, the answer is $\boxed{\text{None of the others}}$ .

By the definition of derivatives

$\displaystyle G'(a)=\lim_{x\rightarrow a}\frac{G(x)-G(a)}{x-a}$

obtaining the derivative $\displaystyle G'(x) = \frac{x}{\sqrt{25 - x^2}}$

Thus, $\displaystyle \lim_{x\rightarrow 1}\frac{G(x)-G(1)}{x-1} = G'(1) = \frac{1}{\sqrt{24}}$