If $f\left(x\right)$ is not continuous at $x=a$, then $f\left(x\right)$ is said to be discontinuous at this point. Figures $1-4$ show the graphs of four functions, two of which are continuous at $x=a$ and two are not.

Classification of Discontinuity Points

All discontinuity points are divided into discontinuities of the first and second kind.

The function $f\left(x\right)$ has a discontinuity of the first kind at $x=a$ if

• There exist left-hand limit $\underset{x\to a-0}{lim}f\left(x\right)$ and right-hand limit $\underset{x\to a+0}{lim}f\left(x\right)$;
• These one-sided limits are finite.

Further there may be the following two options:

• The right-hand limit and the left-hand limit are equal to each other:
  $$\underset{x\to a-0}{lim}f\left(x\right)=\underset{x\to a+0}{lim}f\left(x\right).$$
  Such a point is called a removable discontinuity.
• The right-hand limit and the left-hand limit are unequal:
  $$\underset{x\to a-0}{lim}f\left(x\right)\ne \underset{x\to a+0}{lim}f\left(x\right).$$
  In this case the function $f\left(x\right)$ has a jump discontinuity.

The function $f\left(x\right)$ is said to have a discontinuity of the second kind (or a nonremovable or essential discontinuity) at $x=a$, if at least one of the one-sided limits either does not exist or is infinite.

Solved Problems

Click or tap a problem to see the solution.

Solution.

The given function is not defined at $x=-1$ and $x=1$. Hence, this function has discontinuities at $x=\pm 1$. To determine the type of the discontinuities, we find the one-sided limits:

Since the left-side limit at $x=-1$ is infinity, we have an essential discontinuity at this point.

Similarly, the right-side limit at $x=1$ is infinity. Hence, here we also have an essential discontinuity.

Solution.

Obviously, the function is not defined at $x=0$. As $\sin x$ is continuous at every $x$, then the initial function $f\left(x\right)=\frac{\sin x}{x}$ is also continuous for all $x$ except the point $x=0.$

Since $\underset{x\to 0}{lim}\frac{\sin x}{x}=1,$ the function has a removable discontinuity at this point. We can construct the new function

which is continuous at every real $x.$