# Prove $\lim_{x\to0}\cos(\frac{1}{x})$ does not exist

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.

• What is your $L$? Dec 20, 2013 at 20:40
• @Robert : please explain what "$L$" is in your question (not in a comment) Dec 21, 2013 at 0:16
• Technically the limit is “-1 to 1” inclusive as the limit must be between those vales as Wolfram Alpha sometimes says. Jul 29, 2021 at 20:59

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.
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 $<2/10$.
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 $L$.
Hint $\mathbf{\#2}$: Plot (Computed by Wolfram Mathematica):