Prove the limit of $\cos(1/x)$ does not exist as $x \to 0$, by using Epsilon and Delta.
I was trapped by this question at least for two hours. I negated the definition of limit first, and then try to prove $$\lim_{x \to 0}\cos\left(\frac1x\right)$$ not exist.
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$\begingroup$Have you looked at the graph of this function? If you have, it should be intuitively obvious that no limit exists: the function oscillates wildly as $x \rightarrow 0$, approaching no single limit. It's not immediately clear how this helps you prove it though.
A handy tool to have is sequential continuity: if $f$ is continuous at a point $a$, then given any $a_n \rightarrow a$, we have $f(a_n) \rightarrow f(a)$ (and vice-versa, but we don't need that). A slightly different way to put that is this: if $\lim_{x\rightarrow a} f(x) = L$, then $\lim_{n\rightarrow \infty} f(a_n) = L$ for all $a_n \rightarrow a$.
This can be a handy tool to disprove the existence of a limit, because if we can find two sequences $a_n$ and $b_n$ that both tend to $a$, but for which $f(a_n)$ and $f(b_n)$ don't tend to the same number, then no such limit can exist. This is handy to encapsulate the oscillations of the given function: there are points arbitrarily close to $0$ that take the function value $1$, but also points arbitrarily close to $0$ that take the function value $-1$. This would imply that the limit is both $1$ and $-1$!
Let $a_n = \frac{1}{2n\pi}$ and $b_n = \frac{1}{(2n + 1) \pi}$. Note that the function applied to $a_n$ produces $1$ constantly, and applied to $b_n$ produces $-1$ constantly. Thus $f(a_n) \rightarrow 1$ and $f(b_n) \rightarrow -1$. Thus the limit cannot exist.
$\endgroup$ 2 $\begingroup$Assume the limit is some number $L$, and derive a contradiction. If the limit exists, then for all $\epsilon > 0$, there exists a $\delta > 0$ such that $|\cos(1/x) - L| \leq \epsilon$ for all $x$ satisfying $ 0 < |x| < \delta$. In particular, there exists some $\delta$ for $\epsilon = 1/2$.
Now, let $n$ be a positive integer large enough that $1/n \pi < \delta$. Then we have $$\begin{aligned} 2 &= \left| \cos\left(\frac{1}{1/n \pi}\right) - \cos\left(\frac{1}{1/(n+1)\pi}\right) \right| \\ & \leq \left| \cos\left(\frac{1}{1/n \pi}\right) - L\right| + \left| \cos\left(\frac{1}{1/(n+1) \pi}\right) - L\right| \\ & \leq 2 \epsilon \end{aligned}$$
Where the first inequality follows from the triangle inequality, and the second follows from the fact that $0 < 1/n \pi < \delta$ and $0 < 1/(n+1) \pi < \delta$. Hence $2 \leq 2 \epsilon$, and since $\epsilon = 1/2$ this is absurd, and we have a contradiction.
Note that the intuition behind this is to define the sequence $a_n = 1/(\pi n)$ and notice that $f(a_n)$ constantly oscillates between $1$ and $-1$, and so no matter how small $\delta$ is, there should be points of $a_n$ inside the radius $|x| < \delta$ such that $f(a_n)$ is still oscillating between $1$ and $-1$.
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