Let $$f:R \to R$$   be a differentiable function and $$f\left( 1 \right) = 4.$$   Then the value of $$\mathop {\lim }\limits_{x \to 1} \int\limits_4^{f\left( x \right)} {\frac{{2t}}{{x - 1}}} dt$$     is-

Solution :
We have $$f:R \to R,$$   a differentiable function and $$f\left( 1 \right) = 4$$
NOTE THIS STEP
$$\eqalign{ & \mathop {\lim }\limits_{x \to 1} \int_4^{f\left( x \right)} {\frac{{2t}}{{x - 1}}} dt \cr & = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{{t^2}}}{{x - 1}}} \right]_4^{f\left( x \right)} \cr & = \mathop {\lim }\limits_{x \to 1} \frac{{{{\left( {f\left( x \right)} \right)}^2} - 16}}{{x - 1}} \cr & = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 4}}{{x - 1}}.\mathop {\lim }\limits_{x \to 1} \left( {f\left( x \right) + 4} \right) \cr & = f'\left( 1 \right).\left( {f\left( 1 \right) + 4} \right) \cr & = 8f'\left( 1 \right)\,\,\,\,\,\,\,\,\,\left[ {{\text{Using }}f\left( 1 \right) = 4} \right] \cr} $$