Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{8^x-4^x-2^x+1}{x^2}$

Answer 1

Given:
$\underset{x\to 0}{lim}\frac{8^x-4^x-2^x+1}{x^2}$

$=\underset{x\to 0}{lim}\frac{2^{3x}-2^{2x}-2^x+1}{x^2}$

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

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

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

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

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

$=\left(\log 2^2\right)\left(\log 2\right)$
$=\left(\log 4\right)\left(\log 2\right)$

Therefore,
$\underset{x\to 0}{lim}\frac{8^x-4^x-2^x+1}{x^2}=\left(\log 4\right)\left(\log 2\right)$