Q. Let $f:R \rightarrow R$ be a differentiable function and $f(1) = 4$. Then, the value of $\lim_{x \rightarrow 1}{\int_{4}^{f(x)}\frac{2t}{x - 1}\ dt}\$is
A) $8f^{'}(1)$
B) $4f^{'}(1)$
C) $2f^{'}(1)$
D) $f^{'}(1)$
Explanation
(a)
$\lim_{x \rightarrow 1}{\int_{4}^{f(x)}{\frac{2t}{x - 1}dt = \lim_{x \rightarrow 1}\frac{\int_{4}^{f(x)}{2t\ dt}}{x - 1}}}$ (using L’ Hospital’s rule)
$= \lim_{x \rightarrow 1}\frac{2f(x) \bullet f^{'}(1)}{1}$
$= 2f(1) \bullet f^{'}(1),$ where $f(1) = 4$
$= 8f^{'}(1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.