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If a function defined by $f(x)=\frac{\left(3^x-1\right)^2}{\sin x \log (1+x)}, x \neq 0$, is continuous at $\mathrm{x}=0$, then $\mathrm{f}(0)=$
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The correct answer is:
$\log 3^2$
Given $f(x)=\frac{\left(3^x-1\right)^2}{\sin x \log (1+x)}, x \neq 0$
Now, $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\left(3^x-1\right)^2}{\sin x \log (1+x)}$
$\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(\frac{3^x-1}{x}\right)^2}{\frac{\sin x}{x} \cdot \frac{\log (1+x)}{x}} \\
& =\frac{(\log 3)^2}{1 \times 1}=(\log 3)^2
\end{aligned}$
Since $f(x)$ is continuous at $x=0$
So $\lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow f(0)=(\log 3)^2$
Now, $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\left(3^x-1\right)^2}{\sin x \log (1+x)}$
$\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(\frac{3^x-1}{x}\right)^2}{\frac{\sin x}{x} \cdot \frac{\log (1+x)}{x}} \\
& =\frac{(\log 3)^2}{1 \times 1}=(\log 3)^2
\end{aligned}$
Since $f(x)$ is continuous at $x=0$
So $\lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow f(0)=(\log 3)^2$
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