STD 11 - 12. limits — JEE STD 11 Science — Question

$ = \mathop {\lim }\limits_{h \to 0} \,\sin \,\left( {\frac{{ - 1}}{h}} \right) = \mathop {\lim }\limits_{h \to 0} \,\, - \sin \frac{1}{h}$

$=$ (finite number lies between $-1$ to $1$)

$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} f(0 + h)$

$ = \mathop {\lim }\limits_{h \to 0} \,\,\sin \left( {\frac{1}{h}} \right)$

$=$ (finite number lies between $0$ to $1$)

$ \because \,\,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {0^ + }} f(x)$

$\therefore$ $\mathop {\lim }\limits_{x \to 0} \sin \left( {\frac{1}{x}} \right)$ does not exist.