MCQ
$\mathop {\lim }\limits_{x \to \infty } \frac{{\sin x}}{x} = $
- A$1$
- ✓$0$
- Cઅસ્તિત્વ નથી
- Dએકપણ નહી.
so that $x \to \infty \,\, \Rightarrow \,y \to 0$
$\therefore \,\mathop {\lim }\limits_{x \to \infty } \,\left( {\frac{{\sin x}}{x}} \right) = \mathop {\lim }\limits_{y \to 0} \,\left( {y.\sin \frac{1}{y}} \right)$
$= \mathop {\lim }\limits_{y \to 0} \,y \times \mathop {\lim }\limits_{y \to 0} \,\sin \frac{1}{y} = 0 \times ... = 0$
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