Question

If G (x) = - $ \sqrt{ 25 - x^2}, \, then \, lim_{ x \to 1} \frac{ G \, (x) - G \, (1)}{ x - 1} $ has the value

• A. $\frac{1}{\sqrt{24}}$
• B. $\frac{1}{5}$
• C. $ - \sqrt{24}$
• D. None of these

Answer 1

Answer: (C)

Given, G (x) = - $ \sqrt{ 25 - x^2}$
$\therefore lim_{ x \to 1} \frac{ G \, (x) - G \, (1)}{ x - 1} = lim_{ x \to 1} \frac{ G \, ; (x) - 0}{ 1 - 0} $
$$ [using L' Hospital's rule]
$$ = G' (1) = $ \frac{1}{\sqrt 24} $
$\Bigg [ \because G (x) = - \sqrt{ 25 - x^2} \Rightarrow G' \, (x) = \frac{ 2x}{ 2 \sqrt{ 25 - x^2} } \Bigg ]$