Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{9^x-2\cdot 6^x+4^x}{x^2}$

Answer 1

Given:
$\underset{x\to 0}{lim}\frac{9^x-2\cdot 6^x+4^x}{x^2}$

$=\underset{x\to 0}{lim}\frac{{\left(3^x\right)}^2-2\cdot (3^x\times 2^x)+{\left(2^x\right)}^2}{x^2}$

$=\underset{x\to 0}{lim}\frac{{(3^x-2^x)}^2}{x^2}$
$[\because (a-b{)}^2=a^2-2ab+b^2]$

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

$={\left[\underset{x\to 0}{lim}\frac{(3^x-1)-(2^x-1)}{x}\right]}^2$

$={[\underset{x\to 0}{lim}\frac{3^x-1}{x}-\underset{x\to 0}{lim}\frac{(2^x-1)}{x}]}^2$

$={[\underset{x\to 0}{lim}\frac{3^x-1}{x}-\underset{x\to 0}{lim}\frac{(2^x-1)}{x}]}^2$
$\because \underset{x\to 0}{lim}[\left(\frac{a^x-1}{x}\right)=\log a]$

$=[\log 3-\log 2{]}^2$
$={\left[\log \frac{3}{2}\right]}^2$

Therefore,
$\underset{x\to 0}{lim}\frac{9^x-{2.6}^x+4^x}{x^2}={\left[\log \frac{3}{2}\right]}^2$