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$\lim _{x \rightarrow 0}\left(\frac{3^{x}-1}{x}\right)$ is equal to
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The correct answer is:
$\log 3$
$\lim _{x \rightarrow 0}\left(\frac{3^{x}-1}{x}\right) \quad\left(\frac{0}{0}\right.$ form $)$
Using L' Hospital's rule, $=\lim _{x \rightarrow 0} \frac{3^{x} \log 3-0}{1}=3^{0} \log 3$
Using L' Hospital's rule, $=\lim _{x \rightarrow 0} \frac{3^{x} \log 3-0}{1}=3^{0} \log 3$
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