MCQ
$\mathop {\lim }\limits_{x \to 0} \sin \left( {\frac{1}{x}} \right)$ is
- A$0$
- B$1$
- C$-1$
- ✓Does not exist
$ = \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.
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