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My initial thought would have been yes, but my professor's solution to proving that $\lim_{x \rightarrow0}(2x\sin(1/x)-\cos(1/x))$ doesn't exist has me thinking otherwise.

Professor's solution:

Assume for a contradiction that there exists $\lim_{x \rightarrow0}(2x\sin(1/x)-\cos(1/x))=l$ for some $l\in\mathbb{R}.$

Notice that $\cos(1/x)=(\cos(1/x)-2x\sin(1/x))+2x\sin(1/x).$

Then by our assumption, the fact that $\lim_{x\rightarrow0}2x\sin(1/x)=0$ (proof by Sandwich Theorem omitted) and using the Algebra of Limits we have that $$\lim_{x\rightarrow0}(\cos(1/x)-2x\sin(1/x))+2x\sin(1/x) \\=\lim_{x\rightarrow 0}(\cos(1/x)-2x\sin(1/x))+\lim_{x\rightarrow0}2x\sin(1/x)$$

But observe that $\lim_{x\rightarrow 0}\cos(1/x)$ does not exist (proof using sequences omitted). Hence we have a contradiction.

Ok so I have to admit that I don't even see where the contradiction is nor why we had to go through such a long process, just to end up with needing to show that the limit doesn't exist because $\lim_{x\rightarrow0}\cos(1/x)$ doesn't exist. Couldn't we just have done that from the start?

Alex.F
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    "Couldn't we just have done that from the start?" Okay, and how would you do that? –  Nov 20 '16 at 20:56
  • Well, I agree with you. $\lim{x\to0} \cos(1/x)$ doesn't exist, so the whole limit doesn't exist. Perhaps you are missing part of the original question? – Tim Thayer Nov 20 '16 at 21:00
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    The limit as $x$ tends to $0$ of $1=(x+1)/x-1/x$ exists and yet... – Guest Nov 20 '16 at 21:00
  • @ZacharySelk By using the sequences bit that the professor ended up using anyway. That is, coming up with $(a_n)$ and $(b_n)$ such that neither is equal to 0 for any $n\in\mathbb{N}$ and such that $\cos(1/a_n)$ and $\cos(1/b_n)$ tend to different things. – Alex.F Nov 20 '16 at 21:02
  • @Alex.F: That would prove that $\lim_{x\to 0}\cos(1/x)$ does not exist, but how would that prove the original limit does not exist? – Eric Wofsey Nov 20 '16 at 21:03
  • @Guest Sorry I don't quite understand. Is that $\lim_{x\rightarrow 0}\frac{x(x+1)}{(x-1)}$? Where does the 1 come from? – Alex.F Nov 20 '16 at 21:08
  • @EricWofsey I thought like Tim said that if one part doesn't exist then the whole limit doesn't exist? – Alex.F Nov 20 '16 at 21:09
  • I agree with the OP : the set of convergent sequences is a vector space. As $u_n=x\sin\frac1x$ has a limit and $v_n=\cos\frac1x$ is divergent, their sum can't be convergent. No need to add a pompous and unnatural argument by contradiction. – Nicolas FRANCOIS Nov 20 '16 at 21:11
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    @Alex.F That is $\displaystyle\lim_{x\to0}\left(\frac{x+1}{x}-\frac{1}{x}\right)$. This is $\displaystyle\lim_{x\to0}1$ which exists and equals $1$. Yet, the second term does not tend to a limit as $x\to0$ $\left(\displaystyle\lim_{x\to0}\frac{1}{x}\text{ does not exist}\right)$. You seem to think that this last fact only would imply that the first limit doesn't exist, which is false in general. – Guest Nov 20 '16 at 21:14
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    If the limit of the summands exists, then the limit of the sum exists. The converse is false. – MathematicsStudent1122 Nov 20 '16 at 21:16
  • @Guest Ah I see, that does make sense. How come Tim and Nicolas are suggesting otherwise though? Also, could you help me to see where the contradiction in the professor's solution is please? – Alex.F Nov 20 '16 at 21:23
  • @NicolasFRANCOIS: And saying "the set of convergent sequences is a vector space" is less "pompous"? The argument by contradiction here is literally the exact same argument that you wrote, just with the details you're skipping over filled in (why, exactly, does the set of convergent sequences being a vector space imply that a sum of a convergent sequence and a divergent sequence must diverge?). – Eric Wofsey Nov 20 '16 at 21:24
  • @EricWofsey : Because if $u_n$ converges and $u_n+v_n$ converges leads to $v_n=(u_n+v_n)-u_n$ converges, as you mentioned in your answer. – Nicolas FRANCOIS Nov 20 '16 at 21:58

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The contradiction comes from the following theorem:

If $\lim_{x\to a}f(x)$ and $\lim_{x\to a} g(x)$ both exist, then $\lim_{x\to a}(f(x)+g(x))$ exists (and is equal to $\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$).

This theorem is being applied with $f(x)=\cos(1/x)-2x\sin(1/x)$ and $g(x)=2x\sin(1/x)$. You have assumed that $\lim_{x\to 0}f(x)$ exists, and you have proved that $\lim_{x\to 0}g(x)$ exists. The theorem then tells you that $\lim_{x\to 0}(f(x)+g(x))=\lim_{x\to0}\cos(1/x)$ exists. Since it doesn't exist, this is a contradiction.

Note that it is absolutely essential to prove that $\lim_{x\to 0}g(x)$ exists here. In particular, the following statement which it seems you intend to use (with $h(x)=\cos(1/x)$ and $g(x)=2x\sin(1/x)$) is not true in general:

(FALSE) If $\lim_{x\to a}h(x)$ does not exist, then $\lim_{x\to a}(h(x)-g(x))$ does not exist.

For instance, $\lim_{x\to 0}\frac{1}{x}$ does not exist, but $\lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{x}\right)$ does exist since $\frac{1}{x}-\frac{1}{x}=0$ for all $x\neq 0$.

The FALSE statement above, however, is true if $\lim_{x\to a}g(x)$ exists. The proof is exactly the argument given above: define $f(x)=h(x)-g(x)$, and suppose for a contradiction that $\lim_{x\to a} f(x)$ does exist. Then $\lim_{x\to a}(f(x)+g(x))$ would exist, but this is $\lim_{x\to a} h(x)$ which we know does not exist.

Eric Wofsey
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  • Many thanks for the fantastic explanation. I hope it's okay if I state my understanding just to check I'm now on the right lines. Define $\cos(1/x)=(\cos(1/x)-2x\sin(1/x))+2x\sin(1/x)$. Assume for a contradiction that $\lim_{x\rightarrow0}(\cos(1/x)-2x\sin(1/x))$ exists and we certainly know that $\lim_{x\rightarrow0}2x\sin(1/x)$ exists. So $\lim_{x\rightarrow0}(\cos(1/x)-2x\sin(1/x)+2x\sin(1/x))$ should exist and be equal to $\lim_{x\rightarrow0}\cos(1/x)$ but this last limit doesn’t exist, hence contradiction? – Alex.F Nov 20 '16 at 22:02
  • Yes, that's right. (Though it doesn't make sense to say "Define" the way you did.) – Eric Wofsey Nov 20 '16 at 22:03
  • Nailed it. I think the misconception is that "does not exist" is some magic quantity that annihilates everything else. As in, "does not exist plus anything equals does not exist." – Matthew Leingang Nov 20 '16 at 22:36