Question

please answer this question​

Answer 1

Given Expression,

$f(x) = - \sqrt{25 - {x}^{2} }$

We have to find the value of :

$\displaystyle \lim_{x \longrightarrow \: 1} \: \: \dfrac{f(x) - f(1)}{x - 1}$

Putting x = 0 in the above expression, we obtain an indeterminate form.

Using L'Hospital Rule,

$\displaystyle \lim_{x \longrightarrow a} \dfrac{f(x)}{g(x)} = \lim_{x \longrightarrow a} \dfrac{f'(x)}{g'(x)}$

Now,

$\displaystyle \lim_{x \longrightarrow \: 1} \: \: \dfrac{d( - \sqrt{25 - {x}^{2} } )}{dx} \\ \\ \implies - \displaystyle \lim_{x \longrightarrow \: 1} \: \: \dfrac{d(\sqrt{25 - {x}^{2} } )}{dx} \\ \\ \implies - \displaystyle \lim_{x \longrightarrow \: 1} \: \: \dfrac{1}{2 \sqrt{25 - {x}^{2} } } \times \dfrac{d( - {x}^{2} )}{dx} \\ \\ \implies \displaystyle \lim_{x \longrightarrow \: 1} \: \: \dfrac{2x}{2 \sqrt{25 - {x}^{2} } } \\ \\ \implies \displaystyle \lim_{x \longrightarrow \: 1} \: \: \dfrac{x}{\sqrt{25 - {x}^{2} } } \\ \\ \implies \: \dfrac{1}{ \sqrt{25 - 1} } \\ \\ \implies \: \dfrac{1}{ \sqrt{24} }$

Thus,

$\displaystyle \lim_{x \longrightarrow \: 1} \: \: \dfrac{f(x) - f(1)}{x - 1} = \dfrac{1}{ \sqrt{24} }$