Question

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

Answer 1

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

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

$=\frac{2}{3}.\frac{log3}{log2}[\because \underset{y\to 0}{lim}\left(\frac{a^y-1}{y}\right)=loga]$

$=\frac{2log3}{3log2}=\frac{log\left(3^2\right)}{log\left(2^3\right)}=\frac{log9}{log8}$