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  4. The limit lim limits x arrow ∞ x[ tan -1((x+1/x+2))- tan -1((x/x+2))] is equal to

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Q. The limit $\lim\limits _{x \rightarrow \infty} x\left[\tan ^{-1}\left(\frac{x+1}{x+2}\right)-\tan ^{-1}\left(\frac{x}{x+2}\right)\right]$ is equal to

Inverse Trigonometric Functions

Solution:

$\displaystyle\lim_{x\to\infty} x\left[\tan ^{-1}\left(\frac{x+1}{x+2}\right)-\tan ^{-1}\left(\frac{x}{x+2}\right)\right]$
$=\displaystyle\lim_{x\to\infty}x \tan ^{-1}\left(\frac{\frac{x+1}{x+2}-\frac{x}{x+2}}{1+\frac{x+1}{x+2} \cdot \frac{x}{x+2}}\right)$
$=\displaystyle\lim_{x\to\infty} x \tan ^{-1}\left(\frac{x+2}{2 x^{2}+5 x+4}\right)$
$=\displaystyle\lim_{x\to\infty} x\left(\frac{\tan ^{-1}\left(\frac{x+2}{2 x^{2}+5 x+4}\right)}{\frac{x+2}{2 x^{2}+5 x+4}}\right) \times \frac{x(x+2)}{2 x^{2}+5 x+4}=1 \times \frac{1}{2}=\frac{1}{2}$