Graph of a Sequence of Functions

Question

So, I looked up this question from G.H. Hardy's A Course of Pure Mathematics and found one examination question from the Cambridge Mathematical Tripos and it has baffled me ever since. I am supposed to sketch

$\lim_{n\to\infty}\dfrac{x^{2n}\sin{(\pi x/2)}+x^2}{x^{2n}+1}$

I have found out that this approaches $\sin(\pi x/2)$ (not sure whether am I right or not) and the graphs look like the image attached below but I am told that there is one discontinuity. So, how do I sketch it correctly?

The possible graphs of the sequence of function

Answer 1

We have that $$f_n(x):=\dfrac{x^{2n}\sin{(\pi x/2)}+x^2}{x^{2n}+1}=\sin(\pi x/2)+\dfrac{x^2-\sin{(\pi x/2)}}{x^{2n}+1}.$$ Moreover, $$\lim_{n\to \infty}x^{2n}=\left\{\begin{array}{ll} 0 & \text{if $|x|<1$, } \\ 1 & \text{if $|x|=1$, } \\ +\infty & \text{if $|x|>1$, } \end{array}\right. \quad\Rightarrow\quad \lim_{n\to \infty}\frac{1}{x^{2n}+1}=\left\{\begin{array}{ll} 1 & \text{if $|x|<1$, } \\ 1/2 & \text{if $|x|=1$, } \\ 0 & \text{if $|x|>1$. } \end{array}\right. $$ Therefore $$\lim_{n\to \infty}f_n(x)=\left\{\begin{array}{ll} \sin(\pi x/2) & \text{if $|x|>1$, } \\ 1 & \text{if $x=1$, } \\ 0 & \text{if $x=-1$, } \\ x^2 & \text{if $|x|<1$, } \end{array}\right. .$$

Answer 2

\begin{align*} \lim \limits_{n \rightarrow \infty} \frac{x^{2n}\sin(\pi x/2)+x^2}{x^{2n}+1} &= \lim \limits_{n \rightarrow \infty} \frac{\sin(\pi x/2)+x^{2-2n}}{1+x^{-2n}}\\ &=\lim \limits_{n \rightarrow \infty} \frac{\sin(\pi x/2)+0}{1+0} \end{align*}