If $G\left( x \right) = \sqrt {25 - {x^2}} $, then $\mathop {\lim }\limits_{x \to 1} \frac{{G\left( x \right) - G\left( 1 \right)}}{{x - 1}}$ has the value

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

G\left( x \right) = \sqrt {25 - {x^2}}

, then

\lim\limits_{x \to 1} \frac{{G\left( x \right) - G\left( 1 \right)}}{{x - 1}}

has the value

View answer and explanation

Correct option: A

\frac{1}{{\sqrt {24} }}

Option D (

- 1/5

) is correct because the limit represents the derivative of G(x) evaluated at x=1.

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