Explanation
We have,
\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{3}^{2x}}-{{2}^{3x}}}{x}
Applying L’ Hospital rule and we get,
\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{3}^{2x}}\log 3\times2-{{2}^{3x}}\log 2\times(3)}{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 \dfrac{9}{8}
Hence, this is the answer.
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