Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{8^x-2^x}{x}$

Answer 1

Given:
$\underset{x\to 0}{lim}\frac{8^x-2^x}{x}$

$=\underset{x\to 0}{lim}\frac{8^x-2^x-1+1}{x}$

$=\underset{x\to 0}{lim}\frac{(8^x-1)-(2^x-1)}{x}$

$=\underset{x\to 0}{lim}[\left(\frac{8^x-1}{x}\right)-\left(\frac{2^x-1}{x}\right)]$

$[\because \underset{x\to 0}{lim}\left(\frac{a^x-1}{x}\right)=\log a]$

$=\log 8-\log 2$
$=\log \frac{8}{2}=\log 4[\because \log M-\log N=\log \frac{M}{N}]$

Therefore,
$\underset{x\to 0}{lim}\frac{8^x-2^x}{x}=\log 4$