## Continuity – Limits and Continuity

Posted: 12th February 2013 by seanmathmodelguy in Lectures

#### Continuity

A function $$f(x)$$ is said to be continuous at a point $$a$$ in its domain if the following three properties hold.

• $$\displaystyle \lim_{x \to a} f(x)$$ exists. This takes three steps to show in itself.
• $$f(a)$$ has to exist,
• $$\displaystyle \lim_{x \to a} f(x) = f(a)$$.

Continuity connects the behaviour of a function in a neighbourhood of a point with the value at the point.

If the domain of the function is bounded say $$[a,b]$$ then each end point of the interval can only be approached in one way. The left end point is at $$x = a$$ so $$f(x)$$ is said to be continuous at the left end point ‘$$a$$’ if $$\displaystyle \lim_{x \to a^+} f(x) = f(a)$$. In a similar fashion, $$f(x)$$ is said to be continuous at the right end point ‘$$b$$’ if $$\displaystyle \lim_{x \to b^-} f(x) = f(b)$$.

###### Types of Discontinuities

Since there are only a few ways that the limit of a function cannot exist at a point there are few ways that a function can fail to be continuous. These are classified into four types.
1. Jump Discontinuity (also known as a simple discontinuity)
2. Removable Discontinuity
3. Infinite Discontinuity
4. Oscillatory Discontinuity

An explicit example of each type of discontinuity follows next.

###### Examples

1. Jump Discontinuity:

Is $$f(x) = \begin{cases} \displaystyle x-1, & 1 \le x \le 2 \\ -1, & -2 \le x < 1 \end{cases}$$ continuous at $$x = 1$$? Where is $$f(x)$$ continuous?

To answer this we go back to the definition. By computing $$L^+ = 0$$ and $$L^- = -1$$ (for $$x = 1$$) we see that they are not equal and consequently, $$\displaystyle \lim_{x \to 1} f(x)$$ DNE. Therefore $$f(x)$$ is not continuous at $$x=1$$. As to where $$f(x)$$ is continuous, this is everywhere else in the domain $$(-2,1) \cup (1,2)$$. For the endpoints, in this case we say that $$f(x)$$ is continuous from the right at $$x=-2$$ and continuous from the left at $$x=2$$.
A graph of the function appears to the right.

2. Removable Discontinuity:
Consider $$g(x) = \begin{cases} \displaystyle x, & x \ne 2 \\ 5, & x = 2. \end{cases}$$ Is $$g(x)$$ continuous at $$x = 2$$? No because even though $$\displaystyle \lim_{x\to 2}g(x) = 2$$ exists, $$\displaystyle \lim_{x\to 2}g(x) \ne g(2) = 5$$. Since this function can be made continuous by redefining it at $$x=2$$, we call this type of discontinuity removable.

To illustrate how a function can be fixed, notice that $$f_1(x) = \displaystyle\frac{\sin x}{x}$$ is not continuous for all $$x\in {\Bbb R}$$ since $$f_1(0)$$ DNE. However, $$f_2(x) = \begin{cases} \displaystyle \frac{\sin x}{x}, & x \ne 0 \\[3mm] 1, & x = 0 \end{cases}$$ is continuous for all $$x\in {\Bbb R}$$.

3. Infinite Discontinuity:
Consider the function $$y = 1/x$$ on any domain that includes $$x=0$$. Since the function becomes unbounded continuity fails at $$x=0$$ since the limit does not exist there. The domain of the function is very important since the same function on a different domain ($$\displaystyle y = 1/x, 1 \le x \le 2$$), does not have an infinite discontinuity because it does not become unbounded on the given domain.

4. Oscillatory Discontinuity:
This type of discontinuity occurs when a function oscillates too much, as in the case of $$y = \sin(1/x)$$. As $$x\to 0$$, $$f(x)$$ does not approach a single value.

We leave it as an exercise to the student to show that
$$f(x) = \begin{cases} \displaystyle x\sin\left(\frac{1}{x}\right), & x \ne 0 \\[3mm] 0, & x = 0 \end{cases}$$ is a continuous function for all $$x\in {\Bbb R}$$.

1. Kiru Sengal says:

🙂 There are many other ways a function can be discontinuous.

The list of discontinuities should be sufficient for nice functions (i.e., “elementary” functions aka closed form expressions involving polynomials, trig, log, etc.).