Q. The value of $f(0)$ so that $\frac{( - e^{x} + 2^{x})}{x}$ may be continuous at $x = 0$ is
A) $\log\left( \frac{1}{2} \right)$
B) 0
C) 4
D) $- 1 + \log 2$
Explanation
(d)
Since, $f(x)$ is continuous at $x = 0$
$\therefore\ \ \ \ \lim_{x \rightarrow 0}{f(x)} = f(0)$
$\Rightarrow \ \lim_{x \rightarrow 0}\frac{- e^{x} + 2^{x}}{x} = f(0)$
$\Rightarrow \ \ \lim_{x \rightarrow 0}\frac{- e^{x} + 2^{x}\log 2}{1} = f(0)$ [by L ‘Hospital’s rule]
$\Rightarrow f(0) = - 1 + \log 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.