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Using the definition of a limit, find the limit $$\lim_{x\to 0} x^2\cos\left(\frac1x\right)$$ or prove it does not exist.

I did an example similar to this in class where we figured what the trig function was bounded by and then determined delta was equal to epsilon. However, I do not understand what was done and do not know how to go about doing this problem.

Ben Grossmann
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user72195
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  • I suggest that you start by stating the definition of a limit. If you add this to your original question, we can show you how to use it as a step-by-step guide to solve this problem. Also include the example from class and ask about the specific parts that you don't understand. – Code-Guru Jun 27 '13 at 18:19
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    @user72195 It might be a good idea to ask a question with your notes regarding the example given in class and ask what you didn't understand. – Git Gud Jun 27 '13 at 18:19

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The idea here is this: we know that $x^2\rightarrow0$ as $x\rightarrow0$; we also know that no matter what $x$ happens to be, $$ -1\leq\cos\left(\frac{1}{x}\right)\leq 1.\tag{1} $$ So, no matter what the cosine term does, it cannot "overrule" the tendency of $x^2$ to take us to 0.

To use the definition: let $\epsilon>0$ be given. We want to show that if $x$ is close enough to 0, then $x^2\cos(\frac{1}{x})$ is within $\epsilon$ of 0 - in other words, that $\lvert x^2\cos(\frac{1}{x})\rvert<\epsilon$. But, by the inequalities in (1), $\lvert\cos(\frac{1}{x})\rvert\leq 1$. See if you can use this inequality, along with a proof that $x^2\rightarrow0$ as $x\rightarrow0$, to finish the problem.

Hint: $\delta$ here will be completely determined by the $x^2$ term after you apply the inequality.

Nick Peterson
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  • I am getting caught on the proof from the limit of x^2 I have: let E>0 and choose a delta = ? such that [x^2-0]<d. Now [x^2-0]=X^2 but I do not know where else to go – user72195 Jun 27 '13 at 18:33
  • Maybe this will help: note that $x^2=\lvert x\rvert^2$. And $\lvert x\rvert$ is what we can make small. So, how small do you need to make $\lvert x\rvert$ in order to make $\lvert x\rvert ^2<\epsilon$? – Nick Peterson Jun 27 '13 at 18:35
  • wouldnt it also have to be less than or equal to 1? – user72195 Jun 27 '13 at 18:40
  • That only guarantees that $\lvert x\rvert^2\leq 1\cdot1=1$. Let me put it this way: if $\lvert x\rvert<\delta$, then $\lvert x\rvert^2<\delta^2$. We want to choose $\delta>0$ so that $\delta^2<\epsilon$... if we do that, then $\lvert x\rvert<\delta$ implies $\lvert x\rvert^2<\delta^2<\epsilon$. So, how do you un-do the square root? – Nick Peterson Jun 27 '13 at 18:42
  • then [x]< E but what does that mean? – user72195 Jun 27 '13 at 18:48
  • Not $\lvert x\rvert<\epsilon$... $\lvert x^2\rvert<\epsilon$. Which implies that $\lvert x^2\cos(\frac{1}{x})-0\lvert\leq\lvert x^2\rvert<\epsilon$, which is exactly what you are trying to show. – Nick Peterson Jun 27 '13 at 18:50
  • I hope that this has helped; but, I'm afraid that I can't offer too much more in the way of hints without just writing down a solution to your homework for you (and I'm not prepared to do that). Might I suggest making an appointment with your instructor to go over the problem? They can much more easily draw a picture to help explain it than I can in a forum like this. – Nick Peterson Jun 27 '13 at 18:51
  • that is what I meant, sorry, I mixed up the two expressions I had ([x]^2<E and [x]<square root of E) – user72195 Jun 27 '13 at 18:51
  • Right. So, if $\delta<\sqrt{\epsilon}$, then you can follow through this line of reasoning to finish the proof. – Nick Peterson Jun 27 '13 at 18:52
  • and yes, this has been a help. I just struggle with figuring out what to choose for delta and figuring out where exactly I need to begin problems – user72195 Jun 27 '13 at 18:53
  • Well, that just comes with practice! Keep plugging away; ask for help when you don't understand; and you'll get there. – Nick Peterson Jun 27 '13 at 18:53
  • Why is it that [x] < d? – user72195 Jun 27 '13 at 18:57
  • You are assuming $\lvert x\rvert<\delta$. The entire point here is to show that you can find a $\delta>0$ such that, for any $x$ with $\lvert x-0\vert<\delta$, it follows that $\lvert x^2\cos(\frac{1}{x})-0\rvert<\epsilon$. – Nick Peterson Jun 27 '13 at 18:58
  • @user72195 I hope this helped. If it did: please, consider accepting my answer! – Nick Peterson Jun 27 '13 at 20:46