Question

${lim}_{x\to 0}\frac{3^x-2^x}{4^x-3^x}$ is equal to

• A. 1
• B. -1
• C. 0
• D. none of these

Answer 1

The correct option is D none of these

$\begin{array}{c}{lt}_{x\to 0}\frac{3^x-2^x}{4^x-3^x}={lt}_{x\to 0}\frac{\frac{\left(3^x-1\right)}{x}-\frac{\left(2^x-1\right)}{x}}{\frac{\left(4^x-1\right)}{x}-\frac{\left(3^x-1\right)}{x}}\\ =\frac{\ln 3-\ln 2}{\ln 4-\ln 3}\\ Hence,theoptionDiscorrectanswer.\end{array}$