Question

$\underset{x\to 0}{lim}\frac{3^{2x}-2^{3x}}{x}$ is equal to

• A. $\log \frac{3}{2}$
• B. $1$
• C. $\log \frac{9}{8}$
• D. $0$

Answer 1

The correct option is C $\log \frac{9}{8}$

We have,

${lim}_{x\to 0}\frac{3^{2x}-2^{3x}}{x}$

Applying L’ Hospital rule and we get,

${lim}_{x\to 0}\frac{3^{2x}\log 3\times 2-2^{3x}\log 2\times \left(3\right)}{1}$

Taking limit and we get,

$=3^{2\times 0}\log 3^2-2^{3\times 0}\log 2^3$

$=\log 9-\log 8$

$=\log \frac{9}{8}$

Hence, this is the answer.