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my attempt:

i have to choose one from option A and option D. option B can be eliminated by taking m=1,n=2. option C can also be eliminated by taking m=4, n=3. plz help from choosing one from A and C. thanks.


$f'(x)=mx^{m-1}sin(1/x^n)-nx^{m-n-1}cos(1/x^n)$ (for $x \neq 0$). The sine and cosine function are continuous; therefore differentiability arises if there are factors $x^a$ with $a \geq 0$. Otherwise the derivative of $f(x)$ would have a jump at $x=0$.

  • $\begingroup$ which option seems correct then? $\endgroup$ – ketan Mar 4 '15 at 19:00
  • $\begingroup$ A and D, because (A): $x^{m-n-1}=x^q$ with $q>0$ and for $m=1$ it must hold (strict inequality!) $n=0$, the 2nd term vanishes. $\endgroup$ – kryomaxim Mar 4 '15 at 19:05
  • $\begingroup$ but only one of them is correct. $\endgroup$ – ketan Mar 4 '15 at 19:06
  • $\begingroup$ (D) is not correct; Ex.: m=0,n=1: first term vanishes and second has the factor $x^0$. $\endgroup$ – kryomaxim Mar 4 '15 at 19:09

I'll assume $n\ge1$, otherwise the problem is trivial.

Use the definition of derivative: the derivative exists if and only if $$ f'(0)=\lim_{x\to 0}x^{m-1}\sin\frac{1}{x^n} $$ exists and is finite.

If $m>1$, this limit exists and is zero, because, for $x\ne0$, $$ -|x^{m-1}|\le x^{m-1}\sin\frac{1}{x^n}\le |x^{m-1}| $$ and the squeeze theorem applies, so $f'(0)=0$. If $m\le1$ the limit does not exist.

No hypothesis whatsoever is needed on $n$. So, with this interpretation, A and D are true.

The case would be much different if the question is “does the function have continuous derivative at $0$?” But “differentiable at $0$” usually means “the function has a derivative at $0$”.

If continuous differentiability is required, then we know that $m>1$ and $f'(0)=0$, in this case. Moreover, for $x\ne0$, $$ f'(x)=mx^{m-1}\sin\frac{1}{x^n}-nx^{m-n-1}\cos\frac{1}{x^n} $$ and this is continuous everywhere, except possibly at $0$.

The limit of this function at $0$ should be $0$. Since $m>1$, we already know that $$ \lim_{x\to0}x^{m-1}\sin\frac{1}{x^n}=0 $$ so we also need that $$ \lim_{x\to0}x^{m-n-1}\cos\frac{1}{x^n}=0 $$ Note that if $m-n-1\le0$, the limit doesn't exist. Instead, if $m-n-1>0$, the limit is $0$ with the same argument as before.

So the condition for continuous differentiability at $0$ is $m>n+1$.

For example, if $m=2$ and $n=1$, the derivative is $$ f'(x)=\begin{cases} 2x\sin\dfrac{1}{x}-\cos\dfrac{1}{x} & \text{if $x\ne0$}\\ 0 & \text{if $x=0$} \end{cases} $$ and this function is not continuous at $0$.

So, if we consider this as the question, we have to choose non differentiability, so either C or D.

  • $\begingroup$ is A correct or D? $\endgroup$ – ketan Mar 8 '15 at 4:38
  • $\begingroup$ @ketan If $m>1$ then the function is differentiable at $0$. Suppose $m>n>0$: then $m>1$ and the function is differentiable at $0$; on the other hand, if $n=0$ the function is differentiable. Hence A is correct. But also D is correct, because for $m=1$ and $n>1$ the function is not differentiable at $0$. $\endgroup$ – egreg Mar 8 '15 at 9:39

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