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If $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \ldots \infty$ and $\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k$, then $12 k=$
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The correct answer is:
6
We have,
$\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k$
$\begin{array}{ll}
\Rightarrow & \lim _{x \rightarrow 0} \frac{(1+x) \log (1+x)-x}{x^2}=k \\
\Rightarrow & \lim _{x \rightarrow 0} \frac{\log (1+x)+1-1}{2 x}-=k
\end{array}$
[using L' Hospital Rule]
$\begin{array}{ll}\Rightarrow & \frac{1}{2} \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=k \\ \Rightarrow & \frac{1}{2} \times 1=k \quad\left[\because \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right] \\ \Rightarrow & k=\frac{1}{2} \quad \therefore \quad 12 k=12 \times \frac{1}{2}=6\end{array}$
$\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k$
$\begin{array}{ll}
\Rightarrow & \lim _{x \rightarrow 0} \frac{(1+x) \log (1+x)-x}{x^2}=k \\
\Rightarrow & \lim _{x \rightarrow 0} \frac{\log (1+x)+1-1}{2 x}-=k
\end{array}$
[using L' Hospital Rule]
$\begin{array}{ll}\Rightarrow & \frac{1}{2} \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=k \\ \Rightarrow & \frac{1}{2} \times 1=k \quad\left[\because \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right] \\ \Rightarrow & k=\frac{1}{2} \quad \therefore \quad 12 k=12 \times \frac{1}{2}=6\end{array}$
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