Correct Answer - Option 3 : 0
Concept:
sin θ can take from - ∞ to + ∞ but sin θ gives - 1 to + 1, i.e. for values of θ from - ∞ to + ∞, sin θ always lies between -1 and 1.
\( - 1 \le \sin θ \le 1\)
sin ∞ gives finite value.
Analysis:
Let,
\(f(x)=\mathop {\lim }\limits_{x \to 0} \left( {x\sin \frac{1}{x}} \right)\)
When x = 0
f(x) = 0 × sin ∞
f(x) = 0 × Finite value
\(f(x)=\mathop {\lim }\limits_{x \to 0} \left( {x\sin \frac{1}{x}} \right)=0\)
