lim _{x rightarrow 0} left( frac{3^{x}-1}{x} | vedclass

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Question

(C) The given limit is $\lim _{x \rightarrow 0} \left( \frac{3^{x}-1}{x} \right)$,which is in the $\left( \frac{0}{0} \right)$ indeterminate form.Appl

Vedclass · Accepted Answer

$\log 3$

Vedclass · Answer

(C) The given limit is $\lim _{x ightarrow 0} \left( \frac{3^{x}-1}{x} ight)$,which is in the $\left( \frac{0}{0} ight)$ indeterminate form.Applying $L'	ext{Hospital's rule}$ by differentiating the numerator and the denominator with respect to $x$:$= \lim _{x ightarrow 0} \frac{\frac{d}{dx}(3^{x}-1)}{\frac{d}{dx}(x)}$$= \lim _{x ightarrow 0} \frac{3^{x} \log 3}{1}$$= 3^{0} \log 3 = 1 \cdot \log 3 = \log 3$.

$\lim _{x \rightarrow 0} \left( \frac{3^{x}-1}{x} \right)$ is equal to

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