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Rational Function Limits

Value of rational functions, and cases when denominator equals zero.

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Rationalization to Find Limits
Some limits cannot be evaluated directly by substitution and no factors immediately cancel. In these situations there is another algebraic technique to try called rationalization. With rationalization, you make the numerator and the denominator of an expression rational by using the properties of conjugate pairs.

How do you evaluate the following limit using rationalization?

\begin{align*} \lim \limits_{x \to 16} \frac{\sqrt{x}-4}{x-16}\end{align*}limx16x4x16

Using Rationalization to Find Limits

Rationalization generally means to multiply a rational function by a clever form of one in order to eliminate radical symbols or imaginary numbers in the denominator. Rationalization is also a technique used to evaluate limits in order to avoid having a zero in the denominator when you substitute.

Tod do this, you will use the properties of conjugates.

Conjugates can be used to simplify expressions with a radical in the denominator:

\begin{align*}\frac{5}{1+ \sqrt{3}}=\frac{5}{\left ( 1+ \sqrt{3} \right )} \cdot \frac{\left ( 1- \sqrt{3} \right )}{\left ( 1- \sqrt{3} \right )}=\frac{5-5 \sqrt{3}}{1-3}=\frac{5-5 \sqrt{3}}{-2}\end{align*}51+3=5(1+3)(13)(13)=55313=5532

Conjugates can be used to simplify complex numbers with \begin{align*}i\end{align*}i in the denominator:

\begin{align*}\frac{4}{2+3i}=\frac{4}{(2+3i)} \cdot \frac{(2-3i)}{(2-3i)}=\frac{8-12i}{4+9}=\frac{8-12i}{13}\end{align*}42+3i=4(2+3i)(23i)(23i)=812i4+9=812i13

Here, they can be used to transform an expression in a limit problem that does not immediately factor to one that does immediately factor. 

\begin{align*}\lim \limits_{x \to 16} \frac{\left ( \sqrt{x}-4 \right )}{(x-16)} \cdot \frac{\left ( \sqrt{x}+4 \right )}{\left ( \sqrt{x}+4 \right )}=\lim \limits_{x \to 16} \frac{(x-16)}{(x-16)\left ( \sqrt{x}+4 \right )}\end{align*}limx16(x4)(x16)(x+4)(x+4)=limx16(x16)(x16)(x+4)

Now you can cancel the common factors in the numerator and denominator and use substitution to finish evaluating the limit.

The rationalizing technique works because when you algebraically manipulate the expression in the limit to an equivalent expression, the resulting limit will be the same. Sometimes you must do a variety of different algebraic manipulations in order avoid a zero in the denominator when using the substitution method.

      

Examples

Example 1

In order to evaluate the limit of the following rational expression, you need to multiply by a clever form of 1 so that when you substitute there is no longer a zero factor in the denominator.

\begin{align*}\lim \limits_{x \to 16}\frac{\sqrt{x}-4}{x-16}&=\lim \limits_{x \to 16}\frac{\left(\sqrt{x}-4\right)}{(x-16)} \cdot \frac{\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)} \\ &=\lim \limits_{x \to 16}\frac{(x-16)}{(x-16)\left(\sqrt{x}+4\right)} \\ &=\lim \limits_{x \to 16}\left(\sqrt{x}+4\right) \\ &=4+4\\ &=8\end{align*}limx16x4x16=limx16(x4)(x16)(x+4)(x+4)=limx16(x16)(x16)(x+4)=limx16(x+4)=4+4=8

Example 2

Evaluate the following limit: \begin{align*}\lim_{x \to 3}\frac{x^2-9}{\sqrt{x}-\sqrt{3}}\end{align*}limx3x29x3.

\begin{align*}\lim_{x \to 3}\frac{(x-3)(x+3)}{\left(\sqrt{x}-\sqrt{3}\right)} \cdot \frac{\left(\sqrt{x}+\sqrt{3}\right)}{\left(\sqrt{x}+\sqrt{3}\right)}&=\lim_{x \to 3}\frac{(x-3)(x+3)\left(\sqrt{x}+\sqrt{3}\right)}{(x-3)} \\ &=\lim_{x \to 3}(x+3)\left(\sqrt{x}+\sqrt{3}\right) \\ &=6 \cdot 2\sqrt{3} \\ &=12 \sqrt{3}\end{align*}limx3(x3)(x+3)(x3)(x+3)(x+3)=limx3(x3)(x+3)(x+3)(x3)=limx3(x+3)(x+3)=623=123

Example 3

Evaluate the following limit: \begin{align*}\lim_{x \to 7}\frac{\sqrt{x+2}-3}{x-7}\end{align*}limx7x+23x7.

\begin{align*}\lim_{x \to 7}\frac{\sqrt{x+2}-3}{x-7}&=\lim_{x \to 7}\frac{\left(\sqrt{x+2}-3\right)}{(x-7)} \cdot \frac{\left(\sqrt{x+2}+3\right)}{\left(\sqrt{x+2}+3\right)} \\ &=\lim_{x \to 7}\frac{(x+2-9)}{(x-7) \cdot \left(\sqrt{x+2}+3\right)} \\ &=\lim_{x \to 7}\frac{(x-7)}{(x-7) \cdot \left(\sqrt{x+2}+3\right)} \\ &=\lim_{x \to 7}\frac{1}{\left(\sqrt{x+2}+3\right)} \\ &=\frac{1}{\sqrt{7+2}+3} \\ &=\frac{1}{6}\end{align*}limx7x+23x7=limx7(x+23)(x7)(x+2+3)(x+2+3)=limx7(x+29)(x7)(x+2+3)=limx7(x7)(x7)(x+2+3)=limx71(x+2+3)=17+2+3=16

Example 4

Evaluate the following limit: \begin{align*}\lim_{x \to 0} \frac{(2+x)^{-1}-2^{-1}}{x}\end{align*}limx0(2+x)121x.

\begin{align*}\lim_{x \to 0}\frac{(2+x)^{-1}-2^{-1}}{x}&=\lim_{x \to 0}\frac{\frac{1}{x+2}-\frac{1}{2}}{x} \cdot \frac{(x+2) \cdot 2}{(x+2) \cdot 2} \\ &=\lim_{x \to 0}\frac{2-(x+2)}{2x(x+2)} \\ &=\lim_{x \to 0}\frac{-x}{2x(x+2)} \\ &=\lim_{x \to 0}\frac{-1}{2(x+2)} \\ &=-\frac{1}{2(0+2)} \\ &=-\frac{1}{4} \end{align*}limx0(2+x)121x=limx01x+212x(x+2)2(x+2)2=limx02(x+2)2x(x+2)=limx0x2x(x+2)=limx012(x+2)=12(0+2)=14

Example 5

Evaluate the following limit: \begin{align*}\lim_{x \to 0}\left( \frac{3}{x\sqrt{9-x}}-\frac{1}{x} \right)\end{align*}

\begin{align*}\lim_{x \to 0}\left(\frac{3}{x \sqrt{9-x}}-\frac{1}{x} \right) &=\lim_{x \to 0}\left(\frac{3}{x \sqrt{9-x}}-\frac{\sqrt{9-x}}{x \sqrt{9-x}} \right) \\ &=\lim_{x \to 0}\left(\frac{3- \sqrt{9-x}}{x \sqrt{9-x}} \right) \\ &=\lim_{x \to 0}\left(\frac{\left(3- \sqrt{9-x}\right)}{x \sqrt{9-x}} \cdot \frac{\left(3+ \sqrt{9-x}\right)}{\left(3+ \sqrt{9-x}\right)} \right) \\ &=\lim_{x \to 0}\left(\frac{9-(9-x)}{x \sqrt{9-x}} \right) \\ &=\lim_{x \to 0}\frac{x}{x \sqrt{9-x}} \\ &=\lim_{x \to 0}\frac{1}{\sqrt{9-x}} \\ &=\frac{1}{\sqrt{9-0}} \\ &=\frac{1}{3}\end{align*}

Review

Evaluate the following limits:

1. \begin{align*}\lim_{x \to 9}\frac{\sqrt{x}-3}{x-9}\end{align*}

2. \begin{align*}\lim_{x \to 4}\frac{\sqrt{x}-2}{x-4}\end{align*}

3. \begin{align*}\lim_{x \to 1}\frac{\sqrt{x+3}-2}{x-1}\end{align*}

4. \begin{align*}\lim_{x \to 0}\frac{\sqrt{x+3}-\sqrt{3}}{x}\end{align*}

5. \begin{align*}\lim_{x \to 4}\frac{\sqrt{3x+4}-x}{4-x}\end{align*}

6. \begin{align*}\lim_{x \to 0}\frac{2-\sqrt{x+4}}{x}\end{align*}

7. \begin{align*}\lim_{x \to 0}\frac{\sqrt{x+7}-\sqrt{7}}{x}\end{align*}

8. \begin{align*}\lim_{x \to 16}\frac{16-x}{4-\sqrt{x}}\end{align*}

9. \begin{align*}\lim_{x \to 0}\frac{x^2}{\sqrt{x^2+12}-\sqrt{12}}\end{align*}

10. \begin{align*}\lim_{x \to 2}\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2}\end{align*}

11. \begin{align*}\lim_{x \to 1}\frac{1-\sqrt{x}}{1-x}\end{align*}

12. \begin{align*} \lim_{x \to \frac{1}{9}} \frac{9x-1}{3\sqrt{x}-1}\end{align*}

13. \begin{align*} \lim_{x \to 4} \frac{4x^2-64}{2\sqrt{x}-4}\end{align*}

14. \begin{align*} \lim_{x \to 9}\frac{9x^2-90x+81}{9-3\sqrt{x}}\end{align*}

15. When given a limit to evaluate, how do you know when to use the rationalization technique? What will the function look like?

Review (Answers)

To see the Review answers, open this PDF file and look for section 14.5. 

Vocabulary

Conjugates

Conjugates are pairs of binomials that are equal aside from inverse operations between them, e.g. (3 + 2x) and (3 - 2x).

Continuous

Continuity for a point exists when the left and right sided limits match the function evaluated at that point. For a function to be continuous, the function must be continuous at every single point in an unbroken domain.

limit

A limit is the value that the output of a function approaches as the input of the function approaches a given value.

rationalization

Rationalization generally means to multiply a rational function by a clever form of one in order to eliminate radical symbols or imaginary numbers in the denominator. Rationalization is also a technique used to evaluate limits in order to avoid having a zero in the denominator when you substitute.

theorem

A theorem is a statement that can be proven true using postulates, definitions, and other theorems that have already been proven.

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