Question:
The value of $
\lim_{x \to \infty}\left(\frac{x^2 sin \left(\frac{1}{x}\right)-x}
{1-|x|}\right)$ is
The value of $
\lim_{x \to \infty}\left(\frac{x^2 sin \left(\frac{1}{x}\right)-x}
{1-|x|}\right)$ is
Updated On: Jul 7, 2022
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The Correct Option is A
Solution and Explanation
$\lim_{x\to\infty} \left[\frac{x^{2} \sin\left(\frac{1}{x}\right) -x}{1-\left|x\right|}\right]$
= $\lim_{x\to\infty}\left[\frac{x^{2}\sin\left(\frac{1}{x}\right) -x}{1-x}\right] = \lim_{x\to\infty} \frac{\frac{\sin\left(x^{-1}\right)-1}{x^{-1}}}{x^{-1} -1}$
= $\lim_{y\to0 } \frac{\frac{\sin\,y}{y} -1}{y-1} = \frac{1-1}{0-1}=0$
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Concepts Used:
Limits of Trigonometric Functions
Assume a is any number in the general domain of the corresponding trigonometric function, then we can explain the following limits.
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:
