Search any question & find its solution
Question:
Answered & Verified by Expert
If $G(x)=\sqrt{25-x^{2}}$, then what is $\lim _{x \rightarrow 1} \frac{G(x)-G(1)}{x-1}$ equal to?
Options:
Solution:
2771 Upvotes
Verified Answer
The correct answer is:
$-\frac{1}{2 \sqrt{6}}$
$\mathrm{G}(\mathrm{x})=\sqrt{25-\mathrm{x}^{2}}$
Now,
$\lim _{x \rightarrow 1} \frac{G(x)-G(1)}{x-1}$
$\lim _{x \rightarrow 1} \frac{\sqrt{25-x^{2}}-\sqrt{24}}{x-1}$
Applying L'Hospital rule,
$\lim _{x \rightarrow 1} \frac{(-2 x)}{2 \sqrt{25-x^{2}}}=\frac{-2}{2 \sqrt{25-x^{2}}}$
$=\frac{-2}{2 \sqrt{24}}=\frac{-1}{\sqrt{24}}=\frac{-1}{2 \sqrt{6}}$
Option (a) is correct.
Now,
$\lim _{x \rightarrow 1} \frac{G(x)-G(1)}{x-1}$
$\lim _{x \rightarrow 1} \frac{\sqrt{25-x^{2}}-\sqrt{24}}{x-1}$
Applying L'Hospital rule,
$\lim _{x \rightarrow 1} \frac{(-2 x)}{2 \sqrt{25-x^{2}}}=\frac{-2}{2 \sqrt{25-x^{2}}}$
$=\frac{-2}{2 \sqrt{24}}=\frac{-1}{\sqrt{24}}=\frac{-1}{2 \sqrt{6}}$
Option (a) is correct.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.