$$f:\mathbb R \to \mathbb R,\:f\left(x\right)=\lim _{n\to \infty }\frac{1+x^{2n}}{1+x+...+x^{2n}}$$ I have to find the continuity points of this function.
$$\lim _{n\to \infty }\frac{1+x^{2n}}{1+x+...+x^{2n}}=\lim _{n\to \infty }\frac{\left(1+x^{2n}\right)\left(x-1\right)}{x^{2n+1}-1}$$ I have put the condition $x^{2n+1}-1 \neq 0$ but i don't really know it that does make any sense.
The limit exists for all $x$, so we can simply evaluate it at all points and check for continuity.
If $|x| < 1$, the limits of the numerator and the denominator both exist, so the limit of the ratio is just the ratio of the limits. In this case, it's $1$ for the numerator and $1/(1-x)$ for the denominator, so the value of the function is $1-x$. This is clearly a continuous function for $|x| < 1$.
For $|x| > 1$, divide the numerator and denominator by $x^{2n}$. You'll see you'll get the exact same limit as before, but with $x$ replaced by $x^{-1}$. Since $|x^{-1}| < 1$, the argument above applies, and so the value is $1-1/x$. This is also clearly continuous for $|x| > 1$
The tricky part is where they join together at $x = \pm 1$. To be continuous, the limit as $x$ goes to $\pm 1$ must exist, and it must be equal to the function value there. Looking at the above, the functions for $|x| < 1$ and $|x|>1$ both go to $2$ at $x = -1$ and $0$ at $ x = 1$. So the limits exist.
Next we need to evaluate the function at those points. For $x = - 1$, the ratio is always $2$ independent of $n$, so the function value is $2$ there. For $x = 1$, the ratio is $2/(2n+1)$, which goes to zero as $n$ goes to infinity. So the function value there is $0$. We see that these match the limits from the previous paragraph, meaning the function is continuous at $x =\pm 1$.
We conclude that the function is continuous on its entire domain.