Question
Let $$f:R \to R$$ is differentiable function and $$f\left( 1 \right) = 4,$$ then the value of $$\mathop {\lim }\limits_{x \to 1} \int\limits_0^{f\left( x \right)} {\frac{{2t\,dt}}{{x - 1}}} $$ is :
A.
$$8f'\left( 1 \right)$$
B.
$$4f'\left( 1 \right)$$
C.
$$2f'\left( 1 \right)$$
D.
$$f'\left( 1 \right)$$
Answer :
$$8f'\left( 1 \right)$$
Solution :
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} \int\limits_0^{f\left( x \right)} {\frac{{2t}}{{x - 1}}dt} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{1}{{x - 1}}\left[ {{t^2}} \right]_0^{f\left( x \right)} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{{f^2}\left( x \right) - {f^2}\left( 0 \right)}}{{x - 1}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{2f\left( x \right)f'\left( x \right)}}{1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{using L'Hospital Rule}}} \right) \cr
& = 2f\left( 1 \right)f'\left( 1 \right) \cr
& = 8f'\left( 1 \right) \cr} $$