Question

Evaluate:

$\underset{x\to 0}{lim}\frac{2^x-2^{-x}}{x}$

Answer 1

Now,

$\underset{x\to 0}{lim}\frac{2^x-2^{-x}}{x}$

$=\underset{x\to 0}{lim}\frac{2^x-1-2^{-x}+1}{x}$

$=\underset{x\to 0}{lim}\frac{2^x-1}{x}$$+\underset{-x\to 0}{lim}\frac{2^{-x}-1}{-x}$

$={\log }_{e}2+{\log }_{e}2$

$=2\log 2$

$=\log 4$.