2.4 Continuity  (Page 2/16)

The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given point. These examples illustrate situations in which each of the conditions for continuity in the definition succeed or fail.

Determining continuity at a point, condition 1

Using the definition, determine whether the function $f\left(x\right)=(x^2-4)\text{/}(x-2)$ is continuous at $x=2.$ Justify the conclusion.

Let’s begin by trying to calculate $f\left(2\right).$ We can see that $f\left(2\right)=0\text{/}0,$ which is undefined. Therefore, $f\left(x\right)=\frac{x^2-4}{x-2}$ is discontinuous at 2 because $f\left(2\right)$ is undefined. The graph of $f\left(x\right)$ is shown in [link] .

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Determining continuity at a point, condition 2

Using the definition, determine whether the function $f\left(x\right)=\{\begin{array}{ll}-x^2+4& \text{if}x\le 3\hfill \\ 4x-8& \text{if}x>3\hfill \end{array}$ is continuous at $x=3.$ Justify the conclusion.

Let’s begin by trying to calculate $f\left(3\right).$

$f\left(3\right)=-(3^2)+4=\mathrm{-5}.$

Thus, $f\left(3\right)$ is defined. Next, we calculate $\underset{x\to 3}{\text{lim}}f\left(x\right).$ To do this, we must compute $\underset{x\to 3^{-}}{\text{lim}}f\left(x\right)$ and $\underset{x\to 3^{+}}{\text{lim}}f\left(x\right)\text{:}$

$\underset{x\to 3^{-}}{\text{lim}}f\left(x\right)=-(3^2)+4=\mathrm{-5}$

and

$\underset{x\to 3^{+}}{\text{lim}}f\left(x\right)=4\left(3\right)-8=4.$

Therefore, $\underset{x\to 3}{\text{lim}}f\left(x\right)$ does not exist. Thus, $f\left(x\right)$ is not continuous at 3. The graph of $f\left(x\right)$ is shown in [link] .

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By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can state the following theorem.

Proof

Previously, we showed that if $p\left(x\right)$ and $q\left(x\right)$ are polynomials, $\underset{x\to a}{\text{lim}}p\left(x\right)=p\left(a\right)$ for every polynomial $p\left(x\right)$ and $\underset{x\to a}{\text{lim}}\frac{p\left(x\right)}{q\left(x\right)}=\frac{p\left(a\right)}{q\left(a\right)}$ as long as $q\left(a\right)\ne 0.$ Therefore, polynomials and rational functions are continuous on their domains.

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We now apply [link] to determine the points at which a given rational function is continuous.

Types of discontinuities

As we have seen in [link] and [link] , discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a removable discontinuity    is a discontinuity for which there is a hole in the graph, a jump discontinuity    is a noninfinite discontinuity for which the sections of the function do not meet up, and an infinite discontinuity    is a discontinuity located at a vertical asymptote. [link] illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories.