To prove $f(x)=\lim_{n\to \infty}{x^{2n} \over x^{2n}+1 }$ is discontinuous at $x=\pm1$ I came across this question to show that the given $f(x)$ is continuous at all points on $R$ except $x=\pm1$

$$f(x)=\lim_{n\to \infty}{x^{2n} \over x^{2n}+1 }$$

I know that to be continuous at $x=1$,
$$\lim_{x\to 1}f(x)=f(1)$$
I found $f(1)$ to be ${1 \over 2}$,but I am stuck as to how to calculate $\lim_{x\to 1}f(x)$ as there are two variables involved but its a question on a single variable calculus book.
The simplest way is likely to compute the pointwise limit at every $x$.

When $\lvert x\rvert&lt;1$, $x^{2n}\to0$ as $n\to\infty$, so that 
$$
f(x):=\lim_{n\to\infty}\frac{x^{2n}}{x^{2n}+1}=\frac{0}{0+1}=0.
$$

When $\lvert x\rvert &gt;1$, $x^{2n}\to+\infty$ as $n\to\infty$; so, we can write
$$
f(x):=\lim_{n\to\infty}\frac{x^{2n}}{x^{2n}+1}=\lim_{n\to\infty}\frac{1}{1+\frac{1}{x^{2n}}}=\frac{1}{1+0}=1.
$$

Finally, when $x=\pm1$,
$$
f(x):=\lim_{n\to\infty}\frac{1}{1+1}=\frac{1}{2}.
$$

Now, the fact that this function is discontinuous at $x=\pm1$ is clear.
When $x\not=\pm1$ then the limit equals $1$.  Therefore the limit $f(x)$ as $x$ approaches $\pm1$ equals one.   Otherwise at $x=\pm1$, $f(x)=1/2$.  Such a function is not continuous at $x=\pm1$.