Ytukyg

I have tried this by analizing the limits when $x=2{pi}k$ and $x=k+frac{pi}{2}$ when $kin Bbb N$. In the first case, I have written $lim_{kto 0}cos(frac{1}{2{pi}k})=lim_{kto infty}cos(2{pi}k)=1$ and in the second case $lim_{kto 0}cos(frac{1}{k+frac{pi}{2}})=lim_{kto infty}cos(k+frac{pi}{2})=0$. So the limit doesn't exist.
I would like to know if this is reasoning is correct and the if it would be posible to prove the inexistence of this limit with the $epsilon - delta$ definition.
Thanks in advance

The reasoning is correct, that is we construct two subsequence with two different limit to show that the limit doesn't exist since the limit if exists is unique.

However, there are some typos

$$x_k = frac1{2pi k}$$

Similarly, $$y_k = frac1{left(2k + frac12right) pi}$$

Using the $epsilon$-$delta$ definition is not a huge step from what you have. For any proposed limit $L$, choose $epsilon=frac{1}{2}$. If $|L-0|<epsilon$, then $|1-L|geq frac{1}{2}$, foiling your adversary as there is no $delta$ making all values within $epsilon$ of $L$ after a certain point. If $|L-1|<epsilon$, then $|L|>frac12$, and we exceed a distance of $epsilon$ when the value of the function is $0$.

You actually looked at the sequences $a_k=dfrac{1}{2pi k}$ and $b_k=dfrac{2}{pi+4pi k}$. Each of those approaches $0$ as $k$ goes to infinity. Then you can write that $$lim_{ktoinfty}cos(a_k)=lim_{ktoinfty}cos(2pi k)=1$$ and $$lim_{ktoinfty}cos(b_k)=lim_{ktoinfty}cosleft(frac{pi}{2}+2pi kright)=0$$

By Heine we conclude the limit does not exist.

We will prove that the right-hand limit doesn't exist. Observe that $lim_{x to 0^{+}} frac{1}{x}=infty$.

Since $f(x)=cos x$ is a continuous function,using the above relation we get that: $lim_{ytoinfty}f(y)=f(lim_{x to 0^{+}} frac{1}{x})=lim_{x to 0^+}f(frac{1}{x})=lim_{x to 0^+}cosfrac{1}{x}$.

However, $lim_{ytoinfty}f(y)$ doesn't exist because $f$ is a periodic nonconstant function (see the link provided by @FJ.marsan) ,so $lim_{x to 0^+}cosfrac{1}{x}$ doesn't exist.Hence $lim_{x to 0}cosfrac{1}{x}$ doesn't exist.

The sequence with $x=dfrac1{2kpi}$ is

$$1,1,1,1,1,1,cdots$$

while the sequence with $x=dfrac1{2kpi+dfracpi2}$ is

$$0,0,0,0,0,0,cdots$$

For a pure $epsilon-delta$ proof, notice that whatever $delta$, the range of $cosdfrac1x$ is $[-1,1]$, which doesn't shrink.

Non constant periodic functions has no limit at infinity. Take a look at Why do non-constant periodic functions have no limit at infinity?

Yes your idea is correct with the adjustments already suggested, more simply we can consider

$$x_n=frac2{kpi}to 0$$

then

$$cosleft(frac{1}{x_n}right)=cosleft(frac{kpi}{2}right)$$

which has $3$ different limit points.