Q. Let $f(x) = \left\{ \begin{array}{r} \left( \frac{1}{2} \right)\left\{ g(x) + (x) \right\}\sin\ (x),x \geq 1 \\ \sin{x/x},\ \ \ x < 1 \end{array} \right.\$ Where $g(x) = \left\{ \begin{array}{r} 1,\ \ \ if\ x > 0 \\ - 1,\ \ if\ \ \ \ \ \ x < 0 \\ 0,\ \ \ \ if\ \ \ \ x = 0 \end{array} \right.\$Then, $\lim_{x \rightarrow 1}{f(x)}$ is equal to
A) 0
B) 2
C) $\sin 1$
D) None of these
Explanation
(c)
RHL$= \lim_{x \rightarrow 1^{+}}\frac{1}{2}\left\{ g(x) + (x) \right\}\sin x$
$= \lim_{x \rightarrow 1^{+}}\frac{1}{2}\ \left\{ 1 + x \right\}\sin x$
$= \frac{1}{2}.(1 + 1)\sin 1 = \sin 1\$
and LHL$= \lim_{x \rightarrow 1^{-}}\ \frac{\sin x}{x} = \sin 1$
Since, RHL=LHL=$\sin 1$
$\therefore\ \ \ \ \ \ \lim_{x \rightarrow 1}\ \ f(x) = \sin 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.