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Prove $\lim_{x\to0}\cos\left(\frac{1}{x}\right)$ does not exist using $\epsilon$-$\delta$ proof.

I think my professor wants this to be done by letting $L$ be arbitrary. Then have two cases: $L > 0$ and $L < 1$. Then I would pick an epsilon, for all delta, and pick an $x$.

Any help would be appreciated.

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    $\begingroup$ What is your $L$? $\endgroup$ – nbubis Dec 20 '13 at 20:40
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    $\begingroup$ @Robert : please explain what "$L$" is in your question (not in a comment) $\endgroup$ – Stefan Smith Dec 21 '13 at 0:16
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Suppose $\varepsilon=1/10$. If the limit exists, then there exists $\delta>0$ such that if $|x|<\delta$, then $|\cos x-L|<\varepsilon$. But there are numbers $x_1$, $x_2$, both less than $\delta$ in absolute value, for which $\cos x_1=1$ and $\cos x_2=-1$. So $1$ and $-1$ are both within a distance $1/10$ of $L$. From $1$ to $L$ is a distance $<1/10$ and from $L$ to $-1$ is a distance $<1/10$, so the triangle inequality tells us the distance from $1$ to $-1$ is $<2/10$.

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By the definition of limit, for this limit to exist there should exist a $\delta>0$ for any $\epsilon>0$ such that:$$|cos\frac{1}{x}-L|<\epsilon$$whenever,$$|x|<\delta$$ When $x$ takes small values $cos\frac{1}{x}$ fluctuates rapidly between $1$ and $-1$. So if we fix an arbitrarily small value for $\epsilon$, we can always choose an $x$ such that $$|x|<\delta$$But$$|cos\frac{1}{x}-L|>\epsilon$$ which is a contradiction. So the limit does not exist.

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Hint $\bf\#1$: Well, I guess the 'best' way is to use the "Sequential Criterion":

Let $A\subset \Bbb R$ and let $c$ be a $\color{red}{\text{cluster point}}$ of $A$. A function $f:A\to\Bbb R$ has a limit $L=\lim\limits_{x\to c}f(x)$ if for every $\color{blue}{\text{sequence }}\langle x_n\rangle$ that converges to $c$, $\langle f(x_n)\rangle$ converges to $L$.

If you have not studied it (yet), let me know, so I will adjust the answer accordingly.

Hint $\mathbf{\#2}$: Plot (Computed by Wolfram Mathematica):

enter image description here

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