# 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?

$$\ \lim _{x \rightarrow 16} \frac{\sqrt{x}-4}{x-16}$$

# 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:

$$\ \frac{5}{1+\sqrt{3}}=\frac{5}{(1+\sqrt{3})} \cdot \frac{(1-\sqrt{3})}{(1-\sqrt{3})}=\frac{5-5 \sqrt{3}}{1-3}=\frac{5-5 \sqrt{3}}{-2}$$

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

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

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

$$\ \lim _{x \rightarrow 16} \frac{(\sqrt{x}-4)}{(x-16)} \cdot \frac{(\sqrt{x}+4)}{(\sqrt{x}+4)}=\lim _{x \rightarrow 16} \frac{(x-16)}{(x-16)(\sqrt{x}+4)}$$

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.

$$\ \lim _{x \rightarrow 16} \frac{\sqrt{x}-4}{x-16}$$

Solution

\ \begin{aligned} \lim _{x \rightarrow 16} \frac{\sqrt{x}-4}{x-16} &=\lim _{x \rightarrow 16} \frac{(\sqrt{x}-4)}{(x-16)} \cdot \frac{(\sqrt{x}+4)}{(\sqrt{x}+4)} \\ &=\lim _{x \rightarrow 16} \frac{(x-16)}{(x-16)(\sqrt{x}+4)} \\ &=\lim _{x \rightarrow 16}(\sqrt{x}+4) \\ &=4+4 \\ &=8 \end{aligned}

Example 2

Evaluate the following limit: $$\ \lim _{x \rightarrow 3} \frac{x^{2}-9}{\sqrt{x}-\sqrt{3}}$$.

Solution

\ \begin{aligned} \lim _{x \rightarrow 3} \frac{(x-3)(x+3)}{(\sqrt{x}-\sqrt{3})} \cdot \frac{(\sqrt{x}+\sqrt{3})}{(\sqrt{x}+\sqrt{3})} &=\lim _{x \rightarrow 3} \frac{(x-3)(x+3)(\sqrt{x}+\sqrt{3})}{(x-3)} \\ &=\lim _{x \rightarrow 3}(x+3)(\sqrt{x}+\sqrt{3}) \\ &=6 \cdot 2 \sqrt{3} \\ &=12 \sqrt{3} \end{aligned}

Example 3

Evaluate the following limit: $$\ \lim _{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x-7}$$.

Solution

\ \begin{aligned} \lim _{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x-7} &=\lim _{x \rightarrow 7} \frac{(\sqrt{x+2}-3)}{(x-7)} \cdot \frac{(\sqrt{x+2}+3)}{(\sqrt{x+2}+3)} \\ &=\lim _{x \rightarrow 7} \frac{(x+2-9)}{(x-7) \cdot(\sqrt{x+2}+3)} \\ &=\lim _{x \rightarrow 7} \frac{(x-7)}{(x-7) \cdot(\sqrt{x+2}+3)} \\ &=\lim _{x \rightarrow 7} \frac{1}{(\sqrt{x+2}+3)} \\ &=\frac{1}{\sqrt{7+2}+3} \\ &=\frac{1}{6} \end{aligned}

Example 4

Evaluate the following limit: $$\ \lim _{x \rightarrow 0} \frac{(2+x)^{-1}-2^{-1}}{x}$$.

Solution

\ \begin{aligned} \lim _{x \rightarrow 0} \frac{(2+x)^{-1}-2^{-1}}{x} &=\lim _{x \rightarrow 0} \frac{\frac{1}{x+2}-\frac{1}{2}}{x} \cdot \frac{(x+2) \cdot 2}{(x+2) \cdot 2} \\ &=\lim _{x \rightarrow 0} \frac{2-(x+2)}{2 x(x+2)} \\ &=\lim _{x \rightarrow 0} \frac{-x}{2 x(x+2)} \\ &=\lim _{x \rightarrow 0} \frac{-1}{2(x+2)} \\ &=-\frac{1}{2(0+2)} \\ &=-\frac{1}{4} \end{aligned}

Example 5

Evaluate the following limit: $$\ \lim _{x \rightarrow 0}\left(\frac{3}{x \sqrt{9-x}}-\frac{1}{x}\right)$$

Solution

\ \begin{aligned} \lim _{x \rightarrow 0}\left(\frac{3}{x \sqrt{9-x}}-\frac{1}{x}\right) &=\lim _{x \rightarrow 0}\left(\frac{3}{x \sqrt{9-x}}-\frac{\sqrt{9-x}}{x \sqrt{9-x}}\right) \\ &=\lim _{x \rightarrow 0}\left(\frac{3-\sqrt{9-x}}{x \sqrt{9-x}}\right) \\ &=\lim _{x \rightarrow 0}\left(\frac{(3-\sqrt{9-x})}{x \sqrt{9-x}} \cdot \frac{(3+\sqrt{9-x})}{(3+\sqrt{9-x})}\right) \\ &=\lim _{x \rightarrow 0}\left(\frac{9-(9-x)}{x \sqrt{9-x}}\right) \\ &=\lim _{x \rightarrow 0} \frac{x}{x \sqrt{9-x}} \\ &=\lim _{x \rightarrow 0} \frac{1}{\sqrt{9-x}} \\ &=\frac{1}{\sqrt{9-0}} \\ &=\frac{1}{3} \end{aligned}

# Review

Evaluate the following limits:

1. $$\ \lim _{x \rightarrow 9} \frac{\sqrt{x}-3}{x-9}$$
2. $$\ \lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{x-4}$$
3. $$\ \lim _{x \rightarrow 1} \frac{\sqrt{x+3}-2}{x-1}$$
4. $$\ \lim _{x \rightarrow 0} \frac{\sqrt{x+3}-\sqrt{3}}{x}$$
5. $$\ \lim _{x \rightarrow 4} \frac{\sqrt{3 x+4}-x}{4-x}$$
6. $$\ \lim _{x \rightarrow 0} \frac{2-\sqrt{x+4}}{x}$$
7. $$\ \lim _{x \rightarrow 0} \frac{\sqrt{x+7}-\sqrt{7}}{x}$$
8. $$\ \lim _{x \rightarrow 16} \frac{16-x}{4-\sqrt{x}}$$
9. $$\ \lim _{x \rightarrow 0} \frac{x^{2}}{\sqrt{x^{2}+12}-\sqrt{12}}$$
10. $$\ \lim _{x \rightarrow 2} \frac{\sqrt{2 x+5}-\sqrt{x+7}}{x-2}$$
11. $$\ \lim _{x \rightarrow 1} \frac{1-\sqrt{x}}{1-x}$$
12. $$\ \lim _{x \rightarrow \frac{1}{9}} \frac{9 x-1}{3 \sqrt{x}-1}$$
13. $$\ \lim _{x \rightarrow 4} \frac{4 x^{2}-64}{2 \sqrt{x}-4}$$
14. $$\ \lim _{x \rightarrow 9} \frac{9 x^{2}-90 x+81}{9-3 \sqrt{x}}$$
15. When given a limit to evaluate, how do you know when to use the rationalization technique? What will the function look like?