# Prove limit doesn't exist using $\delta$-$\varepsilon$

Prove in the $\delta$-$\varepsilon$ definition that the limit

$$\lim_{x \to \infty} \frac{3}{2+\sin(x)}$$

does not exist.

I know that $\sin x$ gets different values as $x$ approaches infinity but how do I prove it?

• It will be an $M-\epsilon$ argument rather than $\delta-\epsilon$. – Yves Daoust Dec 16 '14 at 17:29

As a hint, here's a formal proof for why $\lim_{x \to \infty} \cos x$ does not exist. Your problem is very similar, as it also involves infinitely many oscillations for large $x$.
Suppose, towards a contradiction, that $\lim_{x \to \infty} \cos x = L$ for some finite constant $L$. By definition, this means that:
For any $\epsilon > 0$, there is some $M > 0$ such that for any $x \in \mathbb R$, if $x > M$, then $|\cos x - L| < \epsilon$.
In particular, consider $\epsilon = 1/2 > 0$. Then we are guaranteed the existence of some $M > 0$ such that if $x > M$, then $|\cos x - L| < 1/2$. Regardless of the value of this $M$, notice that since $\cos x$ is periodic, we can always find some $x_1,x_2$ larger than $M$ such that $\cos x_1 = 1$ and $\cos x_2 = -1$. Indeed, we can always choose: \begin{align*} x_1 &= 2\pi\lceil M \rceil > M \\ x_2 &= \pi + 2\pi\lceil M \rceil > M \\ \end{align*} where here I used the ceiling function to turn $M$ into an integer. Thus, we know that: \begin{align*} |1 - L| = |\cos x_1 - L| &< 1/2 \\ \left|-1 - L\right| = |\cos x_2 - L| &< 1/2 \\ \end{align*} But this system of inequalities has no solution! Indeed, it follows by the triangle inequality that: $$2 = |2| = |(1 - L) - (-1 - L)| \leq |1 - L| + \left|-1 - L\right| < 1/2 + 1/2 = 1$$ a contradiction. So $\lim_{x \to \infty} \cos x$ does not exist, as desired. $~~\blacksquare$
When $x$ goes to infinity, the function oscillates between $1$ and $3$ forever. So there is no way of making the variation interval smaller than $2$.