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If $\mathrm{f}(x)=\left(\frac{2^{x}-1}{1-3^{x}}\right)$, for $\mathrm{x} \neq 0$ is continuous at $x=0$, then $\mathrm{f}(0)=$
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The correct answer is:
$\frac{-(\log 2)}{(\log 3)}$
$f(0)=\lim _{x \rightarrow 0} \frac{2^{x}-1}{-\left(3^{x}-1\right)}=-\frac{\lim _{x \rightarrow 0} \frac{2^{x}-1}{x}}{\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}}=\frac{-\log 2}{\log 3}$
$$
f(0)=\lim _{x \rightarrow 0} \frac{2^{x}-1}{-\left(3^{x}-1\right)}=-\frac{\lim _{x \rightarrow 0} \frac{2^{x}-1}{x}}{\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}}=\frac{-\log 2}{\log 3}
$$
$$
f(0)=\lim _{x \rightarrow 0} \frac{2^{x}-1}{-\left(3^{x}-1\right)}=-\frac{\lim _{x \rightarrow 0} \frac{2^{x}-1}{x}}{\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}}=\frac{-\log 2}{\log 3}
$$
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