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Rational Function Limits

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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?

 \lim \limits_{x \to 16} \frac{\sqrt{x}-4}{x-16}

Watch This

http://youtu.be/ouWAhqeAaik James Sousa: Find a Limit Requiring Rationalizing

Guidance

The properties of conjugates are used in a variety of places in PreCalculus.

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

\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}

Conjugates can be used to simplify complex numbers with i in the denominator:

\frac{4}{2+3i}=\frac{4}{(2+3i)} \cdot \frac{(2-3i)}{(2-3i)}=\frac{8-12i}{4+9}=\frac{8-12i}{13}

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

\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 )}

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.

Example A

Evaluate the following limit: \lim_{x \to 3}\frac{x^2-9}{\sqrt{x}-\sqrt{3}} .

Solution:

\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}

Example B

Evaluate the following limit:  \lim_{x \to 25} \frac{x-25}{\sqrt{x}-5} .

Solution:  

\lim \limits_{x \to 25}\frac{x-25}{\sqrt{x}-5}&=\lim \limits_{x \to 25}\frac{(x-25)}{\left(\sqrt{x}-5\right)} \cdot \frac{\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)} \\&=\lim \limits_{x \to 25} \frac{(x-25)\left(\sqrt{x}+5\right)}{(x-25)} \\&=\lim \limits_{x \to 25}\left(\sqrt{x}+5\right) \\&=\sqrt{25}+5 \\&=10

Example C

Evaluate the following limit: \lim_{x \to 7}\frac{\sqrt{x+2}-3}{x-7} .

Solution:

\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}

Concept Problem Revisited

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.

\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

Vocabulary

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.

Guided Practice

1. Evaluate the following limit: \lim_{x \to 0} \frac{(2+x)^{-1}-2^{-1}}{x} .

2. Evaluate the following limit: \lim_{x \to -3}\frac{\sqrt{x^2-5}-2}{x+3} .

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

Answers:

1.

\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}

2.

\lim_{x \to -3}\frac{\sqrt{x^2-5}-2}{x+3}&=\lim_{x \to -3}\frac{\left(\sqrt{x^2-5}-2\right)}{(x+3)} \cdot \frac{\left(\sqrt{x^2-5}+2\right)}{\left(\sqrt{x^2-5}+2\right)} \\&=\lim_{x \to -3}\frac{(x^2-5-4)}{(x+3)} \\&=\lim_{x \to -3}\frac{(x^2-9)}{(x+3)} \\&=\lim_{x \to -3}\frac{(x-3)(x+3)}{(x+3)} \\&=\lim_{x \to -3}(x-3) \\&=-3-3 \\&=-6


3.

\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}

Practice

Evaluate the following limits:

1. \lim_{x \to 9}\frac{\sqrt{x}-3}{x-9}

2. \lim_{x \to 4}\frac{\sqrt{x}-2}{x-4}

3. \lim_{x \to 1}\frac{\sqrt{x+3}-2}{x-1}

4. \lim_{x \to 0}\frac{\sqrt{x+3}-\sqrt{3}}{x}

5. \lim_{x \to 4}\frac{\sqrt{3x+4}-x}{4-x}

6. \lim_{x \to 0}\frac{2-\sqrt{x+4}}{x}

7. \lim_{x \to 0}\frac{\sqrt{x+7}-\sqrt{7}}{x}

8. \lim_{x \to 16}\frac{16-x}{4-\sqrt{x}}

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

10. \lim_{x \to 2}\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2}

11. \lim_{x \to 1}\frac{1-\sqrt{x}}{1-x}

12.  \lim_{x \to \frac{1}{9}} \frac{9x-1}{3\sqrt{x}-1}

13.  \lim_{x \to 4} \frac{4x^2-64}{2\sqrt{x}-4}

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

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

Vocabulary

Conjugates

Conjugates

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

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

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

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

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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