Continuity

In the examples on limits in this chapter, we often saw that we compute the limit simply by plugging in the value the variable approaches into the function. While limits are concerned with what is happening around that value and not exactly at that value, evaluating the limit at that value gave us the value of the limit. We attributed this to the property of the functions being 'nice'. In this section, we define more formally what we mean by 'nice' functions.

Definition 2.4 A function $f(x)$ is said to be continuous at $x=a$ if

A function $f(x)$ is continuous on the closed interval $[a, b]$ if it is continuous at each point in the interval. Note that this is the two-sided limit which implies that

Consider a function $f(x)$ as shown in Fig. 2.4 for which we want to determine whether $f(x)$ is continuous at $x=x_{1}$ and $x=x_{2}$. For continuity at a point $x=a$, we need to show that both the $\operatorname{limit}^{\lim _{x \rightarrow a}}$ and $f(a)$ exist and are equal. For $x=x_{1}, f\left(x_{1}\right)=y_{2}$ (recall that a solid marker implies that the endpoint is included while a hollow marker indicates that the endpoint is not included). The (two-sided) $\operatorname{limit}^{\lim _{x \rightarrow x_{1}}} f(x)$ does not exist. We conclude therefore that the function is discontinuous at $x=x_{1}$; this discontinuity is referred to as a jump discontinuity which occurs whenever the graph has a break in it. Now, at $x=x_{2}, f\left(x_{2}\right)=y_{2}$ and $\lim _{x \rightarrow x_{2}} f(x)=y_{2}$; it follows that the function is continuous at $x=x_{2}$. From this example we can give the following informal definition of continuity: a function $f(x)$ is continuous if its graph has no breaks in it. Examples are the function $f(x)=x^{2}$ or any polynomial.

As another example, consider the Heaviside function given by Eq. (2.5) with its graph shown in Fig. 2.3. A jump discontinuity occurs at $x=0$ and, while the function is defined at $x=0[f(0)=1]$, the limit $\lim _{x \rightarrow 0} H(x)$ does not hence the function is discontinuous at $x=0$.

Discontinuities do not only occur when there are breaks in the graphs. Consider a function $f(x)$ given by

the function is not defined at $x=0$ and so it cannot be continuous there. The graph of the function given in Eq. (2.32) is shown in Fig. 2.5. While $\sin (1 / x)$ oscillates as $x$ tends to zero, $\lim _{x \rightarrow 0} f(x)=0$ since $f(x)$ is sandwiched between $-|x|$ (shown in Fig. 2.5 in blue) and $|x|$ (shown in Fig. 2.5 in red). To see this,

Figure 2.4: The graph of a function $y=f(x)$ with a jump discontinuity at $x=x_{1}$. Three points are marked: the endpoint $\left(x_{1}, y_{2}\right)$ (black, solid marker) is included in $f(x)$ while the endpoint $\left(x_{1}, y_{1}\right)$ (hollow marker) is not. The red, solid marker at $\left(x_{2}, y_{2}\right)$ represents a point on the graph at which $f$ is continuous. At $\left(x_{1}, y_{1}\right)$ and $\left(x_{1}, y_{2}\right), f$ is discontinuous.

where $|x \sin (1 / x)|=|x||\sin (1 / x)|$ and since $|\sin (1 / x)| \leq 1$, we have

Similarly, we can show that

yielding

It follows that as $x \rightarrow 0, f(x) \rightarrow 0$.

However, considered an amended definition of $f(x)$ from Eq. (2.32) as follows,

this function is continuous for all real $x$ (including $x=0$ ).

Finally, note that the function $f(x)=\sin (1 / x)$ is not continuous at $x=0$ even if we were to define $f(0)=0$, since the limit of $f(x)$ as $x$ tends to zero does not exist.

Figure 2.5: A plot of the function $y=f(x)$ given by Eq. (2.32) (shown in grey) together with the graphs of $y=|x|$ (red) and $y=-|x|$ (blue). The function is not defined at $x=0$ and the $\operatorname{limit}^{\lim _{x \rightarrow 0} f(x) \text { is } 0}$ as the function is sandwiched between $y=|x|$ and $y=-|x|$.