calculus - Prove that $cos(x)$ doesn't have a limit as $x$ approaches infinity.

I've been working on this one for quite a long time now.

I have to prove that $\cos(x)$ has no limit as $x$ approaches infinity.
Let $\epsilon>o$ and M be any number greater than 0, so that for any x>M: $$|\cos(x)-L| < \epsilon$$
I'm not sure how am I to show that for a specific $L$(anywhere in between $-1$ to $1$) I would choose, I could find ε that would contradict $$|\cos(x)-L| < \epsilon$$
For example, if I choose $\epsilon =1/2$ and $L=1/4$ then for $\cos(x)=1$ the expression above is wrong ,thus $1/4$ isn't the limit, but this happens with any L I chose.

Is giving one example of an arbitrarily chosen L enough?
I cannot give 100 examples, now cannot I?

(tia)

Answer

To show that there is no such $L$, you can show that $\cos x$ will always assume points more than $\epsilon$ apart for some $\epsilon > 0$. Take $\epsilon <1$. $\cos 2n\pi = 1$ for all $n\in\Bbb{Z}$, and $\cos\left((2n + 1)\pi\right) = -1$ for all $n\in\Bbb{Z}$, so that no matter what $N$ you choose, you can always find $x = 2n\pi,x' = (2n + 1)\pi > N$ such that $\left|\cos x - \cos x'\right| = 2 > \epsilon$.

Edit (for completeness): You can use the triangle inequality to show that the above implies so such $L$ can work. Do you see it?

! This implies that no $L$ satisfies $\left|\cos x - L\,\right| < \epsilon$ for all $\epsilon > 0$ whenever $x > M_{\epsilon}$ (some $M_{\epsilon}\in\Bbb{R}$) because if it did, we would have
$$\left|\cos 2n\pi - L\,\right| = \left| 1 - L\,\right| < 1/4$$
for $2n\pi > M_{1/4}$,
$$\left|\cos (2n + 1)\pi - L\,\right| = \left|-1 - L\,\right| = \left| 1 + L\,\right| < 1/4$$
for $(2n+1)\pi > M_{1/4}$, so
$$2 = \left|1 - L + 1 + L\,\right| \leq \left| 1 - L\,\right| + \left| 1 + L\,\right|< 2\cdot \left(1/4\right) < 1/2,$$
which is absurd.