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$\lim _{x \rightarrow \infty} x\left(\log \left(1+\frac{x}{2}\right)-\log \frac{x}{2}\right)=$
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Verified Answer
The correct answer is:
2
$$
\begin{aligned}
& \lim _{x \rightarrow \infty} x\left[\log \left(1+\frac{x}{2}\right)-\log \frac{x}{2}\right] \\
& =\lim _{x \rightarrow \infty} x\left[\frac{\log \left(1+\frac{x}{2}\right)}{\frac{x}{2}}\right]\left\{\because \log a-\log b=\log \frac{a}{b}\right\} \\
& =\lim _{x \rightarrow \infty} x\left[\log \left(\frac{2}{x}+1\right)\right]=\lim _{x \rightarrow \infty} \frac{\log \left(1+\frac{2}{x}\right)}{\frac{1}{x}}
\end{aligned}
$$
By $L$ hospital rule
$$
\begin{aligned}
& =\lim _{x \rightarrow \infty} \frac{\frac{1}{\left(1+\frac{2}{x}\right)}\left(-\frac{2}{x^2}\right)}{\left(-1 / x^2\right)} \\
& =\lim _{x \rightarrow \infty}\left(\frac{2}{1+\frac{2}{x}}\right) \\
& =\frac{2}{1+0}=2
\end{aligned}
$$
\begin{aligned}
& \lim _{x \rightarrow \infty} x\left[\log \left(1+\frac{x}{2}\right)-\log \frac{x}{2}\right] \\
& =\lim _{x \rightarrow \infty} x\left[\frac{\log \left(1+\frac{x}{2}\right)}{\frac{x}{2}}\right]\left\{\because \log a-\log b=\log \frac{a}{b}\right\} \\
& =\lim _{x \rightarrow \infty} x\left[\log \left(\frac{2}{x}+1\right)\right]=\lim _{x \rightarrow \infty} \frac{\log \left(1+\frac{2}{x}\right)}{\frac{1}{x}}
\end{aligned}
$$
By $L$ hospital rule
$$
\begin{aligned}
& =\lim _{x \rightarrow \infty} \frac{\frac{1}{\left(1+\frac{2}{x}\right)}\left(-\frac{2}{x^2}\right)}{\left(-1 / x^2\right)} \\
& =\lim _{x \rightarrow \infty}\left(\frac{2}{1+\frac{2}{x}}\right) \\
& =\frac{2}{1+0}=2
\end{aligned}
$$
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