limits of a function (one-sided limits)

If a function $f(x)$  approaches $c$  as its argument $x$  approaches a point $x_0$ , the function $f$  is said to approach the limit $c$ . Now we shall address the following function.
$$f(x)=\left\{ \begin{array}{cccc}  -x^2 & (x<0)& & \\  1& (x=0)& & \\  x^2 & (x>0)& & \end{array} \right.$$
As this function is not continuous, it is not easy for us to understand the limit of the function.

If $x$  approaches $0$  from the left-side hand, $f(x)\rightarrow 0$ . This result is also same in the case of $x$  approaching from the right-side hand. However, as $f(0)=1$  by its own definition, $\lim_{x\rightarrow 0} f(x)\ne f(0)$ .

We shall define two kinds of limit of a function.  The limit of $f(x)$  in the case which $x\rightarrow x_0$  and $x<x_0$   is said to the left hand limit and is written by
$$\lim_{x\rightarrow -x_0}f(x)$$
If the right hand limit, then
$$\lim_{x\rightarrow +x_0}f(x)$$
These are called one-sided limits. "left hand" is equivalent to "below" and "right hand" is equivalent to "above".

If for any $\epsilon>0$ there is a $\delta>0$ such that $|f(x)-c|<\epsilon$  whenever $0<x_0-x <\delta$ , the limit of $f(x)$  is $c$ as $x$ approaches $x_0$ from below. If the case is from above or right hand, the condition becomes $0<x-x_0<\delta$ .

In the above example, both the right hand and left hand limit of $f(x)$  at the point $0$  are not $f(0)$ unfortunately.