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If the function defined by $f(x)=\frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}, x \neq 0$ is continuous at $x=0$, then $6 f(0)$ is equal to
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3
We have, $f(x)=\frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}$
$$
\begin{aligned}
\Rightarrow \quad f(x) & =\frac{(1+x) \log (1+x)}{x^2}-\frac{1}{x} \\
& =\frac{(1+x) \log (1+x)-x}{x^2}
\end{aligned}
$$
$f(x)$ is continuous at $x=0$
$$
\begin{aligned}
\therefore \quad \lim _{x \rightarrow 0} f(x) & =f(0) \\
\lim _{x \rightarrow 0} f(x) & =\lim _{x \rightarrow 0} \frac{(1+x) \log (1+x)-x}{x^2}
\end{aligned}
$$
Apply L-Hospital's rule
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{(1+x)\left(\frac{1}{1+x}\right)+\log (1+x)-1}{2 x} \\
& \lim _{x \rightarrow 0} \frac{\log (1+x)}{2 x}=\frac{1}{2} \\
& \therefore \quad f(0)=\frac{1}{2} \\
& 6 f(0)=6 \times \frac{1}{2}=3
\end{aligned}
$$
$$
\begin{aligned}
\Rightarrow \quad f(x) & =\frac{(1+x) \log (1+x)}{x^2}-\frac{1}{x} \\
& =\frac{(1+x) \log (1+x)-x}{x^2}
\end{aligned}
$$
$f(x)$ is continuous at $x=0$
$$
\begin{aligned}
\therefore \quad \lim _{x \rightarrow 0} f(x) & =f(0) \\
\lim _{x \rightarrow 0} f(x) & =\lim _{x \rightarrow 0} \frac{(1+x) \log (1+x)-x}{x^2}
\end{aligned}
$$
Apply L-Hospital's rule
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{(1+x)\left(\frac{1}{1+x}\right)+\log (1+x)-1}{2 x} \\
& \lim _{x \rightarrow 0} \frac{\log (1+x)}{2 x}=\frac{1}{2} \\
& \therefore \quad f(0)=\frac{1}{2} \\
& 6 f(0)=6 \times \frac{1}{2}=3
\end{aligned}
$$
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