The value of mathop {lim }limits_{x to infty | vedclass

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(A) Let $x = \frac{1}{t}$. As $x 	o \infty$,$t 	o 0^+$.The expression becomes $\mathop {\lim }\limits_{t 	o 0^+} \frac{(\frac{1}{t^2}) \sin(t) - \f

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(A) Let $x = \frac{1}{t}$. As $x 	o \infty$,$t 	o 0^+$.The expression becomes $\mathop {\lim }\limits_{t 	o 0^+} \frac{(\frac{1}{t^2}) \sin(t) - \frac{1}{t}}{1 - |\frac{1}{t}|}$.Since $x 	o \infty$,$t > 0$,so $|\frac{1}{t}| = \frac{1}{t}$.$= \mathop {\lim }\limits_{t 	o 0^+} \frac{\frac{\sin t}{t^2} - \frac{1}{t}}{1 - \frac{1}{t}} = \mathop {\lim }\limits_{t 	o 0^+} \frac{\frac{\sin t - t}{t^2}}{\frac{t - 1}{t}} = \mathop {\lim }\limits_{t 	o 0^+} \frac{\sin t - t}{t(t - 1)}$.Using the Taylor series expansion $\sin t = t - \frac{t^3}{3!} + \dots$,we get:$= \mathop {\lim }\limits_{t 	o 0^+} \frac{(t - \frac{t^3}{6} + \dots) - t}{t^2 - t} = \mathop {\lim }\limits_{t 	o 0^+} \frac{-\frac{t^3}{6}}{t(t - 1)} = \mathop {\lim }\limits_{t 	o 0^+} \frac{-t^2}{6(t - 1)} = \frac{0}{-6} = 0$.

The value of $\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\sin \frac{1}{x} - x}}{{1 - |x|}}$ is

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