Proof For a Jump Discontinuity. Now, I feel like this is a rather trivial problem, but I'm having a hard time finding a way to prove that $f(x) = \frac{1-\cos(x)}{x^2}$ is discontinuous at $x=0.$
Now my first assumption would be to find two different sequences that go to infinity, but approach different points when plugged into $f(x).$ The two sequences that I initially chose are:$$a_n = \pi + 2n\pi$$ $$b_n = 2n\pi$$
Both $a_n,b_n \rightarrow 0$ as $n \rightarrow \infty$
But if I'm correct, it follows that: $$f(a_n)\rightarrow 0,  n \rightarrow \infty $$
$$f(b_n)\rightarrow 0,  n \rightarrow \infty $$
So this implies that the limit does exist, but there is a removable discontinuity at $x=0.$ Am I missing something? I'm having a hard time finding another pair of sequences that would prove discontinuity. 
 A: Our function is not continuous at $x=0$, because it is not defined at $x=0$. However, the discontinuity at $0$ is removable. To show this, we will show that
$$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}.\tag{1}$$
Here is a proof. Suppose that $x$ is close to $0$ but not equal to $0$. Then
$$\frac{1-\cos x}{x^2}=\frac{(1-\cos x)(1+\cos x)}{(1+\cos x)x^2}=\frac{1}{1+\cos x}\left(\frac{\sin x}{x}\right)^2.$$
We have 
$$\lim_{x\to 0}\frac{1}{1+\cos x}=\frac{1}{2}\quad\text{ and}\quad \lim_{x\to 0}\frac{\sin x}{x}=1.$$
The result (1) follows.
So if we define $f(x)$ by $f(x)=\frac{1-\cos x}{x^2}$ when $x\ne 0$, and $f(x)=\frac{1}{2}$ when $x=0$, the function $f(x)$ will be continuous at $x=0$.
At any point $x\ne 0$, our function is continuous, for it is a simple algebraic combination of standard functions known to be continuous.
A: You say what you want to prove is:

$f(x) = \dfrac{1-\cos(x)}{x^2}$ is discontinuous at $x=0.$

and your subject line refers to a "jump discontinuity".
As a quotient of two continuous functions, this function is continuous on its domain, but its domain does not contain $0$ because the denominator is $0$ at that point.  If it is said that the function is discontinuous at that point, that statement should probably be take to rely for whatever validity it has upon the fact that $0$ is a limit point of the domain, and therefore it may be possible for $\lim\limits_{x\to0} f(x)$ to exist.
For a "jump discontinuity" to exist at $0$, one would have to have $\lim\limits_{x\,\downarrow\,0} f(x)$ and $\lim\limits_{x\,\uparrow\,0} f(x)$ both existing and being two different numbers.
In fact, we have
$$
\lim_{x\to0} \frac{1-\cos x}{x^2} = \frac 1 2,
$$
and that can be shown by using L'Hopital's rule or other methods.  Hence there is no jump discotinuity at $0$.
What exists at $0$ is what is sometimes called a "removable discontinuity".  That means $0$ is a limit point of the domain (so $0$ can be approached from within the domain) and $f(0)$ is undefined but $\lim\limits_{x\to0} f(x)$ is defined and is a finite number, so that that $f$ can be made continuous at $0$ by merely assigning $f(0)$ the one value that would make it continuous there, namely the limit, in this case $1/2$.
