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I have a function f, and I have computed the first and second derivatives of it. The first and second derivative have terms x^4 $\sin \frac{1}{x}$ or similar. I am trying to show whether the function is differentiable at x=0, and i don't understand what to do with the $\sin \frac{1}{x}$ , since at x=0 this would be undefined. What does this say, if anything, about f?

This is in my maths homework, and I was instructed to compute the first and second derivatives, and say whether the function is differentiable at x=0. Im struggling because i'm not sure what the second derivative has to do with differentiability - my suspicion is nothing, and I base this on my understanding of what 'differentiable' means, but you never know.

I am reluctant to give the whole function since I want to understand the concept more than just the answer.

Thanks for your help!

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  • $\begingroup$ You don't do anything with "sin(1/x)" as x goes to 0 because the question does not involve that alone. You cannot, unless they happen to be finite, look at parts of a limit to determine what the whole limit is. What you do know is that sin(1/x) always lies between -1 and 1. What does that tell you about x^4sin(1/x)? $\endgroup$ – user247327 Nov 26 '16 at 22:28
  • $\begingroup$ that x^4sin(1/x) lies between negative and positive infinity? im not sure what more this implies.. $\endgroup$ – Learn4life Nov 26 '16 at 23:08

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