Question:
The value of $\displaystyle\lim_{x\to\infty}\left(\frac{\pi}{2} - \tan^{-1} x\right)^{1/x} $ is
The value of $\displaystyle\lim_{x\to\infty}\left(\frac{\pi}{2} - \tan^{-1} x\right)^{1/x} $ is
Updated On: Apr 19, 2024
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The Correct Option is B
Solution and Explanation
Let $y=\displaystyle\lim _{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan ^{-1} x\right)$
Taking log on both sides, we get $\left(\right.$ form $\left.\frac{\infty}{\infty}\right)$
$\log y=\displaystyle\lim _{x \rightarrow \infty} \frac{1}{x} \log \left(\frac{\pi}{2}-\tan ^{-1} x\right)$
$=\displaystyle\lim _{x \rightarrow \infty} \frac{\left(-\frac{1}{1+x^{2}}\right)}{\frac{\pi}{2}-\tan ^{-1} x}\,\,\,\,$ (using L' Hospital's rule)
$=\displaystyle\lim _{x \rightarrow \infty} \frac{\frac{2 x}{\left(1+x^{2}\right)^{2}}}{-\left(\frac{1}{1+x^{2}}\right)} \,\,\,\,$ (using L' Hospital's rule)
$=\displaystyle\lim _{x \rightarrow \infty} \frac{-2 x}{1+x^{2}}=0$
$\Rightarrow y=e^{0}=1$
Taking log on both sides, we get $\left(\right.$ form $\left.\frac{\infty}{\infty}\right)$
$\log y=\displaystyle\lim _{x \rightarrow \infty} \frac{1}{x} \log \left(\frac{\pi}{2}-\tan ^{-1} x\right)$
$=\displaystyle\lim _{x \rightarrow \infty} \frac{\left(-\frac{1}{1+x^{2}}\right)}{\frac{\pi}{2}-\tan ^{-1} x}\,\,\,\,$ (using L' Hospital's rule)
$=\displaystyle\lim _{x \rightarrow \infty} \frac{\frac{2 x}{\left(1+x^{2}\right)^{2}}}{-\left(\frac{1}{1+x^{2}}\right)} \,\,\,\,$ (using L' Hospital's rule)
$=\displaystyle\lim _{x \rightarrow \infty} \frac{-2 x}{1+x^{2}}=0$
$\Rightarrow y=e^{0}=1$
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Concepts Used:
Limits of Trigonometric Functions
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We know that the graphs of the functions y = sin x and y = cos x detain distinct values between -1 and 1 as represented in the above figure. Thus, the function is swinging between the values, so it will be impossible for us to obtain the limit of y = sin x and y = cos x as x tends to ±∞. Hence, the limits of all six trigonometric functions when x tends to ±∞ are tabulated below:
