1

Show that the following function $ f(x) = x sin \frac 1 x$ is bounded but it's derivative isn't. I don't know how to show this formally.

I found the derivative which is $f'(x)=sin \frac 1x - \frac{cos{\frac 1x}}{x} $ but I'm stuck in showing how to properly show this function is bounded. Graphically $f(x)$ looks like is has a lower bound and its derivative also has a bound.

I'd appreciate if you can explain the answer to this question as well so I can understand.

Arthur Lee
  • 51
  • Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Mar 28 '17 at 13:39
  • Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. – 5xum Mar 28 '17 at 13:40
  • sorry, I'm not an avid user of mathstax I just want to understand how the question is done properly. – Arthur Lee Mar 28 '17 at 13:55
  • Well sure, but it's still good to show some of your own work. Your question is much better now. – 5xum Mar 28 '17 at 13:59

2 Answers2

0

Show that this function is continuous and calculate its limits as $x$ goes to $\infty$ and $-\infty$. Since the function is continuous and both limits exists and are real numbers, the function is continuous.

0

Hint: we have $$|x\sin(1/x)|\le |x|$$

Dr. Sonnhard Graubner
  • 95,283