# Give an example of a rational function f(x) that is continuous for all values of x except at x =...

## Question:

Give an example of a rational function {eq}f(x) {/eq} that is continuous for all values of {eq}x {/eq} except at {eq}x =\pm 5 {/eq}, where it has non-removable discontinuities, and at {eq}x = 2 {/eq}, where it has a removable discontinuity. Justify your answer using limits.

## Removable Discontinuities of a Function:

Suppose {eq}f(x) {/eq} is defined on the intervals {eq}(c-r, c) {/eq} and {eq}(c,c+r) {/eq}. We say that {eq}f(x) {/eq} has a removable discontinuity at {eq}x=c {/eq} if the limit {eq}\lim_{x \to c} f(x) {/eq} exists. This is because the function

{eq}\displaystyle g(x)=\begin{cases} f(x),&x \ne c \\ \lim_{x \to c} f(x), &x = c\end{cases} {/eq}

is continuous at {eq}x=c {/eq} and almost the same as the function {eq}f {/eq}: that is, we can make {eq}f {/eq} be continuous at {eq}x=c {/eq} by defining or redefining {eq}f {/eq} only at the single point {eq}x=c {/eq}.

If {eq}f(x) {/eq} is discontinuous at {eq}x=c {/eq} but does not have a removable discontinuity, we say that {eq}f(x) {/eq} has a non-removable discontinuity.

## Answer and Explanation:

A rational function {eq}f(x)=\dfrac{P(x)}{Q(x)} {/eq} will be discontinuous at the zeroes of its {eq}Q(x) {/eq}. So if {eq}R(x) {/eq} has...

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View this answerA rational function {eq}f(x)=\dfrac{P(x)}{Q(x)} {/eq} will be discontinuous at the zeroes of its {eq}Q(x) {/eq}. So if {eq}R(x) {/eq} has discontinuities only at {eq}x=\pm 5 {/eq} and {eq}x=2 {/eq}, then {eq}Q(x) {/eq} should have the zeroes {eq}x=\pm 5 {/eq} and {eq}x=2 {/eq} and no other zeroes. So we'll take {eq}Q(x)=(x-2)(x+5)(x-5) {/eq}. If the discontinuity at {eq}x=2 {/eq} is to be removable, then we want it to be canceled out by the numerator everywhere except at the point {eq}x=2 {/eq}, so we'll take {eq}P(x)=x-2 {/eq}.

That is, we'll take {eq}f(x)=\frac{x-2}{(x-2)(x+5)(x-5)} {/eq}. To verify that this {eq}f {/eq} satisfies the conditions of the problem statement, we'll take limits. First:

{eq}\begin{align*} \lim_{x \to 2}f(x)&=\lim_{x \to 2}\frac{x-2}{(x-2)(x+5)(x-5)}\\ &=\lim_{x \to 2} \frac{1}{(x+5)(x-5)}&\text{(canceling; the limit doesn't depend on the behavior of the function at }x=2\text{)}\\ &=\frac{1}{(2+5)(2-5)}&\text{(because }\frac{1}{(x+5)(x-5)}\text{ is a rational function whose denominator is nonzero at }x=2\text{ and therefore is continuous at }x=2\text{)}\\ &=\frac{1}{7(-3)}\\ &=-\frac{1}{21} \, . \end{align*} {/eq}

Since {eq}\lim_{x \to 2} f(x) {/eq} exists, {eq}f(x) {/eq} has a removable discontinuity at {eq}x=5 {/eq}.

Next:

{eq}\begin{align*} \lim_{x \to 5} f(x)&=\lim_{x \to 5} \frac{x-2}{(x-2)(x+5)(x-5)}\\ &=\frac{5-2}{(5-2)(5+5)}\lim_{x \to 5} \frac{1}{x-5}&\text{(because the function }\frac{x-2}{(x-2)(x+5)}\text{ is rational with nonzero numerator and denominator at }x=5\text{, so it is continuous and nonzero at }x=5\text{)}\\ &=\frac{3}{3(10)}\lim_{x \to 5} \frac{1}{x-5}\\ &=\frac{1}{10}\lim_{x \to 5} \frac{1}{x-5} \, . \end{align*} {/eq}

But {eq}\lim_{x \to 5} \frac{1}{x-5} {/eq} diverges, so {eq}\lim_{x \to 5} f(x) {/eq} must also diverge. Thus {eq}f(x) {/eq} has a non-removable discontinuity at {eq}x=5 {/eq}.