$\lim\limits_{x\to\infty }(\frac{4}{5+\cos x})$ does not have a finite limit using Cauchy definition

$\lim\limits_{x\to\infty }(\frac{4}{5+\cos x})$ prove that the limit is not finite while $x \to \infty$ using Cauchy definition

Need to prove using this definition according to the textbook let $f$ be definied in the interval $(M_0 , \infty)$ , we'll say that $L \in \Bbb R$ is the limit of $f$ at infinity if $\forall \epsilon >0 , \exists M \in \Bbb R >M_0 , \forall x , M<x $ we will get $|f(x)-L|<\epsilon$

so since we need to prove that the limit is NOT infinite I think we need to prove this: if $\exists \epsilon >0 , \forall M \in \Bbb R >M_0 , \exists x , M<x $ we will get $|f(x)-L|\geq \epsilon$ not sure if my negating is right of the definition.

so we have $f(x)=\frac{4}{5+\cos x}$ and we need to prove $\lim\limits_{x\to\infty }(\frac{4}{5+\cos x})\not = L$

I believe we need to separate into 2 cases (I do not know the reason but I saw the lecturer do something similar in a similar question appreciate if someone can explain this part please) because $-1 \leq \cos x \leq 1$ so if $\cos x = 1$ we get $f(x)= \frac{4}{5+\cos x} \geq \frac{4}{6}$ and if $\cos x = -1$ we get $f(x)= \frac{4}{5+\cos x} \leq \frac{4}{4}=1$ so our function is $\frac{4}{6} \leq \frac{4}{5+\cos x} \leq 1$

back to the proof , let $L \in \Bbb R$ we will show that there exists $\epsilon >0$ such that for all $M \in \Bbb R$ there exists $x>M$ such that $|f(x)-L| \geq \epsilon$

we will check for $L=\frac{4}{6}$ and $L \not = \frac{4}{6}$

I got stuck here $|\frac {4}{5+\cos x}- \frac{4}{6}|=|\frac{4-4 \cos x}{30+6 \cos x}|=|\frac{2-2 \cos x}{15+3 \cos x}|\geq \epsilon$

I cannot find a way to get rid of the absolute value and $\cos x$ and I am not sure of my ways as I am tryng to do what the teacher did but it is not much of understanding .. appreciate any help and tips!

Solution 1:

Suppose that limit $L$ exists. For $\epsilon = \dfrac{1}{10}$,$\exists M_1$ such that $x > M_1 \implies |f(x) - L| < \dfrac{1}{10}$. Let $N_1\in \mathbb{N}$ be such that $2\pi N_1 > M_1$. Then if $n > N_1 \implies 2n\pi > 2\pi N_1 > M_1 \implies |f(2n\pi) - L| < \dfrac{1}{10}\implies \left|\dfrac{2}{3}-L\right| = \left|\dfrac{4}{5+\cos(2n\pi)} - L \right| < \dfrac{1}{10}$. Next for $\epsilon = \dfrac{1}{15}, \exists M_2$ such that $x > M_2 \implies |f(x) - L| < \dfrac{1}{15}$. Again, let $N_2 \in \mathbb{N}$ be such that $(2N_2+1)\pi > M_2$. So if $n > N_2 \implies (2n+1)\pi > (2N_2+1)\pi > M_2\implies |f((2n+1)\pi) - L| < \dfrac{1}{15}\implies |1-L| < \dfrac{1}{15}$. Thus if $n > N_1+N_2\implies \dfrac{1}{3} \le |1-L| + \left|\dfrac{2}{3} - L\right| < \dfrac{1}{10} + \dfrac{1}{15} = \dfrac{1}{6}\implies \dfrac{1}{3} < \dfrac{1}{6}$. Contradiction. So the limit $L$ cannot exist.

Solution 2:

Let use notation $f(x) = \frac{4}{5+\cos x}$.Now, for any $M$ you can find values $x_1,x_2 > M$ such that $f(x_1) = 1$ and $f(x_2) = \frac{2}{3}$, and thus $|f(x_1) - f(x_2)| = \frac{1}{3}$. Pick any $0<\epsilon < 0.1$ Now we can prove by the absense of a limit by contradiction. If a limit $L$ exists than we pick any $0<\epsilon < \frac{1}{6}$ such that $\forall x_1,x_2 > M$ we have $|f(x_1) -L| < \epsilon$ and $|f(x_2) -L| < \epsilon$ then $|f(x_1)-f(x_2)| < 2\epsilon$ which contradicts the fact that we can always find $x_1,x_2 > M$ such that $|f(x_1)-f(x_2) = \frac{1}{3}$.

Solution 3:

No periodic function, $f(n)$, has a limit at infinity. For any two elements in its range,e.g. $L_1, L_2$ and an arbitrarily large natural N, There exists some n greater than that so that $f(n)$ equal to one or the other of those two elements. Either way, the difference is greater than $\epsilon =|L_1-L_2|/2$. Since we always have this $n$, $\epsilon$ pair for any $N$, the limit does not exist.