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TS EAMCET 04 May 2019 Shift 2 Solved Paper

Section: Mathematics

Question : 62 of 160

Marks: +1, -0

\lim\limits_{x\to\infty} x\left(\log\left(1+\frac{x}{2}\right)-\log\frac{x}{2}\right)

Solution:

Consider the limit

\lim\limits_{x\to\infty}\left[\log\left(1+\;\frac{x}{2}\right)-\log\;\frac{x}{2}\right]

It is solved as,

\lim\limits_{x\to\infty}\left[\log\left(1+\;\frac{x}{2}\right)-\log\;\frac{x}{2}\right] \;=\lim\limits_{x\to\infty}\left[\log\left(\frac{1+\frac{x}{2}}{\frac{x}{2}}\right)\right]

\;=\lim\limits_{x\to\infty} x\left[\log\left(\frac{2}{x}+1\right)\right]

\;=\lim\limits_{x\to\infty}\left[\frac{\log\left(\frac{2}{x}+1\right)}{\frac{1}{x}}\right]

\;=\lim\limits_{x\to\infty}\left[\frac{\left(1+\frac{2}{x}\right)\left(-\frac{2}{x^{2}}\right)}{-\frac{1}{x^{2}}}\right]

Solve further

\lim\limits_{x\to\infty}\left[\log\left(1+\;\frac{x}{2}\right)-\log\;\frac{x}{2}\right] \;=\lim\limits_{x\to\infty}\left[\frac{2}{1+\frac{2}{x}}\right]

\;=\frac{2}{1+0}

\;=2

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