Question
If \[f\left( x \right) = \left\{ \begin{array}{l}
{x^n}\sin \left( {\frac{1}{{{x^2}}}} \right),\,\,x \ne 0\\
0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0
\end{array} \right.,\left( {n\, \in \,I} \right),\] then :
A.
$$\mathop {\lim }\limits_{x \to 0} f\left( x \right){\text{ exist for }}n > 1$$
B.
$$\mathop {\lim }\limits_{x \to 0} f\left( x \right){\text{ exist for }}n < 0$$
C.
$$\mathop {\lim }\limits_{x \to 0} f\left( x \right){\text{ does not exist for any value of }}n$$
D.
$$\mathop {\lim }\limits_{x \to 0} f\left( x \right){\text{ cannot be determined}}$$
Answer :
$$\mathop {\lim }\limits_{x \to 0} f\left( x \right){\text{ exist for }}n > 1$$
Solution :
$$\eqalign{
& {\text{For }}n > 1, \cr
& \mathop {\lim }\limits_{x \to 0} {x^n}\sin \left( {\frac{1}{{{x^2}}}} \right) = 0 \times \left( {{\text{any value between}} - 1{\text{ and }}1} \right) = 0 \cr
& {\text{For }}n < 0, \cr
& \mathop {\lim }\limits_{x \to 0} {x^n}\sin \left( {\frac{1}{{{x^2}}}} \right) = \infty \times \left( {{\text{any value between}} - 1{\text{ and }}1} \right) = \infty \cr} $$