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)
$\endgroup$ 26 Answers
$\begingroup$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?
$\endgroup$ 4 $\begingroup$! 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.
If you want a more general way. Pick an L between -1 and 1. And an $\epsilon < 1$. Note that the cosine function bijective on a period. Therefore you can find an $x$ such that $|cos(x) - L| >\epsilon. $ Now note that the same holds for $|cos(x+n2\pi) - L|$ with $n$ a natural number.
$\endgroup$ 4 $\begingroup$Suppose the limit is L, and show that it can't be. Hint: $\cos \pi n = (-1)^n$.
$\endgroup$ $\begingroup$Hint:
Observe that $\cos (x)$ is a Periodic function provided $f(x)=\cos x$ is not a constant as Stahl suggests.
$\endgroup$ 7 $\begingroup$Hint: if such limit, say $L$, exists, then any sequence of the form $a_n:=\cos(x_n)$, $n \in \mathbb{N},\ x_n \in dom( \cos x)$ satisfies $$a_n \longrightarrow L.$$ Now, think if this is true for all sequences in the mentioned form.
$\endgroup$ 1 $\begingroup$Let $\epsilon\leq1$, and for arbitrary $N>0$, we pick $n$ such that $x=2\pi$ $n>N$ for some $n$; now, $\cos(x)= 1$, but $\cos(x+\pi)= -1$ and we have a contradiction.
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