Question:
If \(f(x) = \sqrt{25 – x^2}\), then what is \(\lim_{x \to 1} \dfrac{f(x) – f(1)}{x – 1}\) equal to?
यदि \(f(x) = \sqrt{25 – x^2}\) है, तो \(\lim_{x \to 1} \dfrac{f(x) – f(1)}{x – 1}\) किसके बराबर है?
Options:
- \(-\dfrac{1}{4\sqrt{3}}\)
- \(-\dfrac{1}{\sqrt{24}}\)
- \(\dfrac{1}{\sqrt{24}}\)
- \(\dfrac{1}{4\sqrt{3}}\)
- \(-\dfrac{1}{4\sqrt{3}}\)
- \(-\dfrac{1}{\sqrt{24}}\)
- \(\dfrac{1}{\sqrt{24}}\)
- \(\dfrac{1}{4\sqrt{3}}\)
Correct Answer: b)
Explanation:
By definition of derivative: \(\lim_{x \to 1} \dfrac{f(x)-f(1)}{x-1} = f'(1)\).
\(f'(x) = \dfrac{-x}{\sqrt{25-x^2}} \implies f'(1) = -\dfrac{1}{\sqrt{24}}\).
अवकलज की परिभाषा से: \(\lim_{x \to 1} \dfrac{f(x)-f(1)}{x-1} = f'(1) = -\dfrac{1}{\sqrt{24}}\)।