Question

The value of $f(0)$ so that $\frac{\left(-e^{x} +2^{x}\right)}{x}$may be continuous at $x = 0$ is

• A. $log\left(\frac{1}{2}\right)$
• B. 0
• C. 4
• D. - 1 + log 2

Answer 1

Answer: (D)

- 1 + log 2

Answer 2

$f \left(x\right)=\frac{-e^{x} + 2^{x}}{x}$
$=\frac{1}{x}\left[-\left(1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+..\right)+1+\frac{log 2}{1!}x+\frac{\left(bg 2\right)^{2}}{2!}x^{2}+\frac{\left(log 2\right)^{3}}{3!}x^{3}+..\right]$
f \left(x\right)=log 2-1+\frac{x}{2!}\left\\{\left(log 2\right)^{2} -\right\\} +\frac{x^{2}}{3!}\left\\{\left(log^{2}\right)^{3}-1\right\\}+....
Putting $x = 0$, we get
$f\left(0\right) = log 2 - 1 + 0 + 0 + .... = - 1 + log \,2.$