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)

$f\left(x\right)=\frac{-e^x+2^x}{x}$
$=\frac{1}{x}\left[-\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+..\right)+1+\frac{log2}{1!}x+\frac{{\left(bg2\right)}^2}{2!}{x}^2+\frac{{\left(log2\right)}^3}{3!}{x}^3+..\right]$
$f\left(x\right)=log2-1+\frac{x}{2!}\left\{{\left(log2\right)}^2-\right\}<br/>+\frac{x^2}{3!}\left\{{\left(lo{g}^2\right)}^3-1\right\}+....$
Putting $x=0$, we get
$f\left(0\right)=log2-1+0+0+....=-1+log2.$