Question

The value of $f$ at $x=0$ so that function $f\left(x\right)=\frac{2^x-2^{-x}}{x},x\ne 0$, is continuous at $x=0$, is

• A. $0$
• B. $\log 2$
• C. $4$
• D. $\log 4$

Answer 1

The correct option is D $\log 4$

$\underset{x\to 0}{lim}\frac{2^x-2^{-x}}{x}=\underset{x\to 0}{lim}{2}^x\log 2+2^{-x}\log 2$
(by L'Hospital's rule)
$=\log 2+\log 2=\log 4$
Since, the function is continuous at $x=0$.
$\therefore f\left(0\right)=\underset{x\to 0}{lim}\frac{2^x-2^{-x}}{x}$
$\Rightarrow f\left(0\right)=\log 4$