Is it enough to show that $\lim_{x\rightarrow 0}\cos(1/x)$ doesn't exist to show that $\lim_{x \rightarrow0}(2x\sin(1/x)-\cos(1/x))$ doesn't exist? 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?
 A: 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.
