Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{3^{2+x}-9}{x}$

Answer 1

Given:
$\underset{x\to 0}{lim}\frac{3^{2+x}-9}{x}$

$=\underset{x\to 0}{lim}\frac{3^2\times 3^x-9}{x}$
$[\because a^{x+y}=a^x.a^y]$

$=\underset{x\to 0}{lim}\frac{9(3^x-1)}{x}$

$=9\underset{x\to 0}{lim}\frac{(3^x-1)}{x}$

$=9\log 3$ $[\because \underset{x\to 0}{lim}\left(\frac{a^x-1}{x}\right)=\log a]$

Therefore,
$\underset{x\to 0}{lim}\frac{3^{2+x}-9}{x}=9\log 3$