Evaluate: \(\lim\limits_{x \to 0}(\frac{3^{2x}-2^{3x}}{x})\)

1.	Evaluate: \(\lim\limits_{x \to 0}(\frac{3^{2x}-2^{3x}}{x})\)
Answer» \(\lim\limits_{x \to 0}(\frac{3^{2x}-2^{3x}}{x})\) \(=\lim\limits_{x \to 0}\{\frac{(3^{2x}-1)-(2^{3x}-1)}{x}\}\) \(=\lim\limits_{x \to 0}\frac{3^{2x}-1}{x}-\lim\limits_{x \to 0}\frac{2^{3x}-1}{x}\) \(=\lim\limits_{x \to 0}\frac{1}{2}[\frac{3^{2x}-1}{x}]-3.[\lim\limits_{x \to 0}\frac{2^{3x}-1}{x}]\) = 2 ∙ (log 3) − 3(log 2) = log3² − log2³ = log \(\frac{9}{8}\)

Answer»

\(\lim\limits_{x \to 0}(\frac{3^{2x}-2^{3x}}{x})\)

\(=\lim\limits_{x \to 0}\{\frac{(3^{2x}-1)-(2^{3x}-1)}{x}\}\)

\(=\lim\limits_{x \to 0}\frac{3^{2x}-1}{x}-\lim\limits_{x \to 0}\frac{2^{3x}-1}{x}\)

\(=\lim\limits_{x \to 0}\frac{1}{2}[\frac{3^{2x}-1}{x}]-3.[\lim\limits_{x \to 0}\frac{2^{3x}-1}{x}]\)

= 2 ∙ (log 3) − 3(log 2)

= log3² − log2³

= log \(\frac{9}{8}\)

Discussion

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