Correct Answer - Option 1 :
\( - \frac{1}{{\sqrt {24} }}\)
Concept:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
- \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \;\frac{0}{0}\)
- \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \;\frac{\infty }{\infty }\)
Then we can apply L-Hospital Rule ⇔ \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \;\mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\)
Note: We have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = l \ne \frac{0}{0}\) where l is a finite value.
Calculation:
Given: \(f\left( x \right) = \sqrt {25 - {x^2}} \)
We have to find the value of \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}\)
\( \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \;\frac{0}{0}\;\left( {Form} \right)\)
Apply L-Hospital Rule,
\( \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{f'\left( x \right) - 0}}{{1 - 0}} = f'\left( 1 \right)\)
Now, \(f\left( x \right) = \sqrt {25 - {x^2}} \)
\( \Rightarrow {{\rm{f}}^{\rm{'}}}\left( {\rm{x}} \right) = {\rm{\;}}\frac{{1{\rm{\;}} \times \left( {0 - 2{\rm{x}}} \right)}}{{2\sqrt {25 - {{\rm{x}}^2}} }} = {\rm{\;}}\frac{{ - {\rm{x}}}}{{\sqrt {25 - {{\rm{x}}^2}} }}\)
\( \Rightarrow {{\rm{f}}^{\rm{'}}}\left( 1 \right) = {\rm{\;}}\frac{{ - 1}}{{\sqrt {25 - {1^2}} }} = \;\frac{{ - 1}}{{\sqrt {24} }}\)
∴
\(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \;f'\left( 1 \right) = \;\frac{{ - 1}}{{\sqrt {24} }}\)
