Evaluating $ \lim_{x\rightarrow \infty}e^{-x } + 2\cos(3x)$ Find the limit or prove that it does not exist by $\varepsilon-\delta$ approach:
$$ \lim_{x\rightarrow \infty}e^{-x } + 2\cos(3x)$$

Note:I found this question when I was doing exercise from the book Calculus:Early Transcendentals. The book just need me to show it does not exist, but I think it would be interesting to strictly prove it by $\varepsilon-\delta$ language.


Update: This question is not duplicate to the other question at all, since I am asking a strictly proof by $\varepsilon-\delta$ approach here rather than just show it does not exist. I also emphasized the requirement in my Note when I post this question. Although those two questions hold the same functions, but they have different requirements.Hence, It actually totally different from another one.

 A: No limit, since the limit along the sequence of general term $2n\pi$ is $2$ and the limit along the sequence of general term $2n\pi+\pi$ is $-2$. The asymptotics is best described by saying that the limit set is $[-2,2]$ (this is the set of possible limits).
Edit: To prove that the function $f:x\mapsto\mathrm e^{-x}+2\cos(3x)$ has no limit when $x\to+\infty$ using what you call the epsilon-delta approach, one can show:
$$
\forall N\gt0,\exists x\gt N,\exists y\gt N,|f(x)-f(y)|\gt1.
$$
Now that we know that $x$ and $y$ are to be looked for near odd and even multiples of $\pi$, the rest should follow.
A: The first term goes to zero, but the limit of the second it not defined.  However, the limit is bounded between $-2$ and $2$.
A: Existence of a limit $a$ means that $\exists a  \, \forall \varepsilon>0 \, \exists M > 0: \forall x \ge M \, | f(x)-a| < \varepsilon$. Let's negate that: $\forall a \, \exists \varepsilon>0 \, \forall M>0 \, \exists x \ge M: |f(x)-a|\geq \epsilon $. Take $\varepsilon=0.1$ and observe that $\forall x \, \exists z \in (x, x+2\pi): | \cos(z) - \cos(x) | \ge 1$, simply because cosine oscillates. If $\cos x\ge0$, then it must be that $x\in[2k\pi, (2k+1)\pi]$ for $k=\lfloor x/\pi \rfloor/2$, and then you can choose $z=(2k+3/2)\pi$ to get $\cos z=-1$. Then if the limit $a$ existed, fixing $\varepsilon=0.1$, we would have $|{\rm e}^{-M} + 2\cos(3x) - a| < 0.1$. Based on our earlier observation, we can find $z \in (M, M+2\pi/3)$ s.t. $|\cos(3M) - \cos(3z)| \ge 1$. But then ${\rm e}^{-z} + 2 \cos(3z)$ is at least $- {\rm e}^{-M} + {\rm e}^{-z} - 0.1 + 2 > 1.9 - {\rm e}^{-M} \ge 0.9 > 0.1$ away from $a$, QED.
A: Proof: Since the value of $f(x) = e^{-x} + 2\cos(3x)$ oscillate in interval $[e^{-x}-2,~e^{-x}+2]$, then for arbitrary number $N$, I am always able to choose $~x_{1},x_{2}\gt N$, such that $f(x_{1})=0$ and $f(x_{2})=1$. Assume the limit exist as $x$ goes to infinity, then
$$\forall\varepsilon\gt0,\exists N\gt0,~such\space that \space x \gt N,~|f(x)-L| \lt \varepsilon \tag{1}$$  
Let $\varepsilon = \frac{1}{2}$. Because $~x_{1},x_{2}\gt N$，then by $(1)$, we must both have:
\begin{align*}
\ &|f(x_{1}) - L|= |L| \lt \frac{1}{2} \tag{2}
\\&|f(x_{2})-L|=|1-L|\lt \frac{1}{2} \tag{3}
\end{align*}  
(2),(3) inequality cannot simultaneously be true for L. Thus by contradiction, there is no limit of $e^{-x} + 2\cos(3x)$ as $x$ towards infinity.
