A) \[\underset{x\to 0}{\mathop{\lim }}\,f(x)\]exist for\[n>1\]
B) \[f\]is continuous at\[x=\text{0}\]for\[n>1\]
C) \[f\]is differentiable at\[x=\text{0}\]for\[n>1\]
D) None of the above
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,{{x}^{n}}\sin \left( \frac{1}{{{x}^{2}}} \right)=0\] (\[\because \]\[\sin x\]has any value between\[-1\]and 1) Hence,\[f(x)\]is continuous, when\[n>0\]. Now, \[f(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{n}}.\sin \left( \frac{1}{{{h}^{2}}} \right)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,{{h}^{n-1}}.\sin \left( \frac{1}{{{h}^{2}}} \right)=0\] (\[\because \sin x\]has any value between\[-1\]and 1) Hence,\[f(x)\]is differentiable, when \[n>1\].
You need to login to perform this action.
You will be redirected in
3 sec
