﻿ Indeterminate Limits---Rationalizing 0/0 Forms: Examples and interactive practice problems explained step by step

# Indeterminate Limits---Rationalizing $$\frac 0 0$$ Forms

### Quick Overview

For limits with $$\frac 0 0$$ forms that involve square-roots, try rationalizing the numerator (or denominator).

### Examples

##### Example 1

Evaluate: $$\displaystyle \lim_{x\to3}\,% \frac{x-3} {\sqrt{x+22}-5}%$$

Step 1

Confirm that the limit has an indeterminate.

$$\displaystyle \lim_{x\to3}\,% \frac{x-3} {\sqrt{x+22} - 5}% % = \frac{3-3}{\sqrt{3+22}-5} % = \frac 0 {\sqrt{25}-5} % = \frac 0 0 \qquad\mbox{Indeterminate}$$

Step 2

Rationalize the denominator , then divide out the common factors.

\\ \begin{align*} \lim_{x\to3}\,\frac {x-3} {\sqrt{x+22}-5}% % & = \lim_{x\to3}\,% \frac {x-3} {\sqrt{x+22}-5} \cdot \frac {\blue{\sqrt{x+22}+5}} {\blue{\sqrt{x+22}+5}}% && \substack{\Large{\text{Multiply by}\hspace{2mm} \\ \text{the conjugate}}} \\[6pt] % & = \lim_{x\to3}\, \frac {(x-3)(\sqrt{x+22}+5)} {(x+22)-25} \\[6pt] % & = \lim_{x\to3}\, \frac {\red{(x-3)}(\sqrt{x+22}+5)} {\red{x-3}} && \substack{\Large{\text{Divide out}\hspace{10mm} \\ \text{common factors}}} \\[6pt] % & = \lim_{x\to3}\, \frac {\sqrt{x+22}+5} 1 \\[6pt] % & = \lim_{x\to3}\,(\sqrt{x+22}+5) \end{align*} \\

Step 3

Evaluate the simpler limit.

$$\displaystyle \lim_{x\to3}\,(\sqrt{x+22}+5) % = \sqrt{25} + 5 % = 5 + 5 % = 10$$

$$\displaystyle \lim_{x\to3}\,% \frac {x-3} {\sqrt{x+22}-5} % = 10$$

##### Example 2

Evaluate: $$\displaystyle \lim_{x\to13} \frac{\sqrt{x-4} - 3}{x-13}$$

Step 1

Confirm that the limit laws will give you an indeterminate form.

$$\\ \displaystyle \lim_{x\to13}\, \frac {\sqrt{x-4} - 3} {x-13} % = \frac{\sqrt{13-4} - 3} {13-13} % = \frac{\sqrt 9 - 3} 0 % = \frac 0 0% \qquad\mbox{Indeterminate} \\$$

Step 2

Rationalize the numerator. Then divide out the common factors.

\\ \begin{align*} \displaystyle\lim_{x\to13}\,% \frac {\sqrt{x-4} - 3} {x-13}% % & = \displaystyle\lim_{x\to13}\,% \frac {\sqrt{x-4} - 3} {x-13}% \cdot% \frac {\blue{\sqrt{x-4} + 3}} {\blue{\sqrt{x-4} + 3}}% && \substack{\Large{\text{Multiply by}\hspace{1mm} \\ \text{ the conjugate}}} \\[8pt] % & = \displaystyle\lim_{x\to13}\, \frac {(x-4) - 9} {(x-13)(\sqrt{x-4} + 3)} \\[8pt] % & = \displaystyle\lim_{x\to13}\, \frac {\red{x-13}} {\red{(x-13)}(\sqrt{x-4} + 3)}% && \substack{\Large{\text{Divide out}\hspace{10mm} \\ \text{common factors}}} \\[8pt] % & = \displaystyle\lim_{x\to13}\,\frac 1 {\sqrt{x-4}+3} \end{align*} \\

Step 3

Evaluate the simpler limit.

$$\displaystyle\lim_{x\to 13}\,% \frac 1 {\sqrt{x-4} + 3}% = \frac 1 {\sqrt{13-4} + 3}% = \frac 1 {\sqrt 9 +3}% = \frac 1 6$$

$$\displaystyle \lim_{x\to13}\, \frac {\sqrt{x-4} - 3} {x-13} = \frac 1 6$$

### Practice Problems

Step 1

Confirm that the limit laws give you an indeterminate form.

$$\displaystyle\lim_{x\to 4}\,% \frac {x-4} {\sqrt{x+5} - 3} = \frac{4 -4} {\sqrt{4+5}-3} = \frac 0 {\sqrt 9 -3} = \frac 0 0 \qquad\mbox{Indeterminate!}$$

Step 2

Rationalize the denominator . Then divide out the common factors.

\\ \begin{align*} \displaystyle\lim_{x\to 4}\,% \frac {x-4} {\sqrt{x+5} - 3} & = \displaystyle\lim_{x\to 4}\, \frac {x-4} {\sqrt{x+5} - 3} \cdot \frac {\blue{\sqrt{x+5} + 3}} {\blue{\sqrt{x+5} + 3}} && \substack{\Large{\text{Multiply by}\hspace{1mm} \\ \text{the conjugate}}} \\[6pt] % & = \displaystyle\lim_{x\to 4}\, \frac {(x-4)(\sqrt{x+5} + 3)} {(x+5) - 9} \\[6pt] % & = \displaystyle\lim_{x\to 4}\, \frac {\red{(x-4)}(\sqrt{x+5} + 3)} {\red{x - 4}} && \substack{\Large{\text{Divide out}\hspace{10mm} \\ \text{common factors}}} \\[6pt] % & = \displaystyle\lim_{x\to 4}\, \frac{\sqrt{x+5} + 3} 1 \\[6pt] % & = \displaystyle\lim_{x\to 4}\,(\sqrt{x+5} + 3) \end{align*} \\

Step 3

Evaluate the simpler limit.

$$\displaystyle \lim_{x\to 4}\,(\sqrt{x+5} + 3) = \sqrt{4 + 5} + 3 = \sqrt 9 +3 = 6$$

$$\displaystyle \lim_{x\to 4}\, \frac {x-4} {\sqrt{x+5} - 3} = 6$$

Step 1

Confirm the limit laws give you an indeterminate form.

$$\displaystyle\lim_{x\to5}\, \frac {\sqrt{x-1}-2} {x-5} =\frac{\sqrt{5 -1} -2} {5-5} = \frac{\sqrt 4 -2} 0 = \frac 0 0 \qquad\mbox{Indeterminate!}$$

Step 2

Rationalize the numerator. Then divide out the common factors.

\\ \begin{align*} \displaystyle\lim_{x\to5}\, \frac {\sqrt{x-1}-2} {x-5} & = \displaystyle\lim_{x\to5}\, \frac {\sqrt{x-1}-2} {x-5} \cdot \frac {\blue{\sqrt{x-1}+2}} {\blue{\sqrt{x-1}+2}} &&\substack{\Large{\text{Multiply by} \\ \text{the conjugate}}} \\[6pt] % & = \displaystyle\lim_{x\to5}\, \frac {(x-1)-4} {(x-5)(\sqrt{x-1}+2)} \\[6pt] % & = \displaystyle\lim_{x\to5}\, \frac {\red{x-5}} {\red{(x-5)}(\sqrt{x-1}+2)} && \substack{\Large{\text{Divide out the} \\ \text{common factor}}} \\[6pt] % & = \displaystyle\lim_{x\to5}\,\frac 1 {\sqrt{x-1}+2}\\[6pt] \end{align*} \\

Step 3

Evaluate the simpler limit.

$$\displaystyle\lim_{x\to5}\, \frac 1 {\sqrt{x-1}+2} = \frac 1 {\sqrt{5-1} + 2} = \frac 1 {\sqrt 4 + 2} = \frac 1 4$$

$$\displaystyle \lim_{x\to5}\, \frac{\sqrt{x-1}-2} {x-5} = \frac 1 4$$

Step 1

Confirm the limit has an indeterminate form.

$$\displaystyle\lim_{x\to 1}\, \frac {9 - \sqrt{x+80}} {x-1} = \frac{9 - \sqrt{80+1}} {1-1} = \frac{9 - \sqrt{81}} 0 = \frac 0 0 \qquad \mbox{Indeterminate!}$$

Step 2

Rationalize the numerator. Then divide out the common factors.

\\ \begin{align*} \displaystyle\lim_{x\to 1}\, \frac {9 - \sqrt{x+80}} {x-1} & = \displaystyle\lim_{x\to 1}\, \frac {9 - \sqrt{x+80}} {x-1} \cdot \frac {\blue{9 + \sqrt{x+80}}} {\blue{9 + \sqrt{x+80}}} && \substack{\Large{\text{Multiply by}\hspace{2mm} \\ \text{the conjugate}}} \\[6pt] % & = \displaystyle\lim_{x\to 1}\, \frac {81 - (x+80)} {(x-1)(9 + \sqrt{x+80})} \\[6pt] % & = \displaystyle\lim_{x\to 1}\, \frac {81 - x - 80} {(x-1)(9 + \sqrt{x+80})} \\[6pt] % & = \displaystyle\lim_{x\to 1}\, \frac {1 - x} {(x-1)(9 + \sqrt{x+80})} && \substack{\Large{\text{Factor -1 out of}\hspace{10mm} \\ \text{the numerator}\hspace{11mm}}} \\[6pt] % & = \displaystyle\lim_{x\to 1}\, \frac {-\red{(x-1)}} {\red{(x-1)}(9 + \sqrt{x+80})} &&\substack{\Large{\text{Divide out}\hspace{10mm} \\ \text{common factors}}} \\[6pt] % & = \displaystyle\lim_{x\to 1}\,\frac{-1}{9 + \sqrt{x+80}} \end{align*} \\

Step 3

Evaluate the simpler limit.

$$\displaystyle\lim_{x\to 1}\, \frac{-1}{9 + \sqrt{x+80}} = \frac{-1}{9 +\sqrt{1+80}} = -\frac 1 {9+\sqrt{81}} = -\frac 1 {18}$$

$$\displaystyle \lim_{x\to 1}\, \frac {9 - \sqrt{x+80}} {x-1} = -\frac 1 {18}$$

Step 1

Verify the limit has an indeterminate form.

$$\displaystyle \lim_{x\to 6}\, \frac {3x-18} {10-\sqrt{13x+22}} = \frac{3(6)-18} {10-\sqrt{13(6)+22}} = \frac 0 {10-10} = \frac 0 0 \qquad\mbox{Indeterminate!}$$

Step 2

\\ \begin{align*} \small{\lim_{x\to 6}\, \frac {3x-18} {10-\sqrt{13x+22}} } & \small{= \lim_{x\to 6}\, \frac {3x-18} {10-\sqrt{13x+22}} } \cdot \small {\frac {\blue{10+\sqrt{13x+22}}} {\blue{10+\sqrt{13x+22}}} } && \substack{\large{\text{Multiply by}\hspace{2mm} \\ \text{the conjugate}}} \\[6pt] % & \small{ = \lim_{x\to 6}\, \frac {(3x-18)(10+\sqrt{13x+22})} {100-(13x+22)} } \\[6pt] % & \small { = \lim_{x\to 6}\, \frac {3(x-6)(10+\sqrt{13x+22})} {100-13x-22} } \\[6pt] % & \small { = \lim_{x\to 6}\, \frac {3(x-6)(10+\sqrt{13x+22})} {78-13x} } \\[6pt] % & \small { = \lim_{x\to 6}\, \frac {3\blue{(x-6)}(10+\sqrt{13x+22})} {-13\blue{(x-6)}} } && \substack{\large{\text{Divide out}\hspace{10mm} \\ \text{common factors}}} \\[6pt] % & \small { = \lim_{x\to 6}\, \frac{3(10+\sqrt{13x+22})}{-13} } \end{align*} \\

Step 3

Evaluate the simpler limit.

$$\small{ \displaystyle\lim_{x\to 6}\, \frac{3(10+\sqrt{13x+22})}{-13} = -\frac{3(10+\sqrt{13(6)+22})}{13} = -\frac{3(10+\sqrt{100})}{13} = -\frac{3(20)}{13} = -\frac{60}{13}}$$

$$\displaystyle \lim_{x\to 6}\, \frac {3x-18} {10-\sqrt{13x+22}} = -\frac{60}{13}$$

Step 1

Verify the limit has an indeterminate form.

$$\displaystyle\lim_{x\to-8}\, \frac {x^2+9x+8} {\sqrt{x+12}-2} =\frac{(-8)^2+9(-8) + 8} {\sqrt{-8+12}-2} =\frac{64 - 72 + 8} {\sqrt 4 - 2} = \frac 0 0 \quad\mbox{Indeterminate!}$$

Step 2

\\ \begin{align*} \displaystyle\lim_{x\to-8}\, \frac {x^2+9x+8} {\sqrt{x+12}-2} & = \displaystyle\lim_{x\to-8}\, \frac {x^2+9x+8} {\sqrt{x+12}-2} \cdot \frac {\blue{\sqrt{x+12}+2}} {\blue{\sqrt{x+12}+2}} && \substack{\Large{\text{Multiply by}\hspace{2mm} \\ \text{the conjugate}}} \\[6pt] % & = \displaystyle\lim_{x\to-8}\, \frac {(x^2+9x+8)(\sqrt{x+12}+2)} {(x+12)-4} \\[6pt] % & = \displaystyle\lim_{x\to-8}\, \frac{% (x^2+9x+8) (\sqrt{x+12}+2) } {x+12-4} \\[6pt] % & = \displaystyle\lim_{x\to-8}\, \frac{% \blue{(x^2+9x+8)} (\sqrt{x+12}+2) } {x+8} && \substack{\Large{\text{Factor the} \\ \text{numerator}}} \\[6pt] % & = \displaystyle\lim_{x\to-8}\, \frac{% \red{(x+8)} (x+1) (\sqrt{x+12}+2) } {\red{x+8}} && \substack{\Large{\text{Divide out}\hspace{10mm} \\ \text{common factors}}} \\[6pt] % & = \displaystyle\lim_{x\to-8}\, \frac{(x+1)(\sqrt{x+12}+2)}{1}\\[6pt] % & = \displaystyle\lim_{x\to-8}\,(x+1)(\sqrt{x+12}+2) \end{align*} \\

Step 3

Evaluate the simpler limit.

$$\displaystyle\lim_{x\to-8}\,(x+1)(\sqrt{x+12}+2) = (-8+1)(\sqrt{-8+12}+2) = -7(\sqrt 4 + 2) = -28$$

$$\displaystyle \lim_{x\to-8}\, \frac {x^2+9x+8} {\sqrt{x+12}-2} = -28$$

Step 1

Confirm the limit has an indeterminate form.

$$\displaystyle \lim_{x\to 3}\, \frac {\sqrt{x+6} - 3} {\sqrt{x+13} - 4} =\frac{\sqrt{3+6} - 3} {\sqrt{3+13}-4} =\frac{\sqrt 9 -3} {\sqrt{16}-4} = \frac 0 0 \qquad\mbox{Indeterminate!}$$

Step 2

Rationalize the numerator and the denominator. The solution below begins with the denominator.

\\ \begin{align*} &\quad \displaystyle\lim_{x\to 3}\, \frac {\sqrt{x+6} - 3} {\sqrt{x+13} - 4} \\[6pt] & = \displaystyle\lim_{x\to 3}\, \frac {\sqrt{x+6} - 3} {\sqrt{x+13} - 4} \cdot \frac {\blue{\sqrt{x+13}+4}} {\blue{\sqrt{x+13}+4}} && \substack{\Large{\text{Multiply by the conjugate} \\ \text{of the denominator.}\hspace{11mm}}} \\[6pt] % & = \displaystyle\lim_{x\to 3}\, \frac{% (\sqrt{x+13}+4) (\sqrt{x+6} - 3) } {(x+13) - 16} \\[6pt] % & = \displaystyle\lim_{x\to 3}\, \frac{% (\sqrt{x+13}+4) (\sqrt{x+6} - 3) } {x - 3} \\[6pt] % & = \displaystyle\lim_{x\to 3}\, \frac{% (\sqrt{x+13}+4) (\sqrt{x+6} - 3) } {x - 3} \cdot \frac {\red{\sqrt{x+6}+3}} {\red{\sqrt{x+6}+3}} && \substack{\Large{\text{Multiply by the conjugate} \\ \text{of the original numerator.}}} \\[6pt] % & = \displaystyle\lim_{x\to 3}\, \frac{% (\sqrt{x+13}+4) ((x+6) - 9) } {% (x - 3) (\sqrt{x+6}+3) } \\[6pt] % & = \displaystyle\lim_{x\to 3}\, \frac{% (\sqrt{x+13}+4) \red{(x - 3)} } {% \red{(x - 3)} (\sqrt{x+6}+3) } && \substack{\Large{\text{Divide out the}\hspace{2mm} \\ \text{common factor.}}} \\[6pt] % & = \displaystyle\lim_{x\to 3}\ \frac {\sqrt{x+13}+4} {\sqrt{x+6}+3} \end{align*} \\

Step 3

Evaluate the simpler limit.

$$\displaystyle \lim_{x\to 3}\, \frac {\sqrt{x+13}+4} {\sqrt{x+6}+3} =\frac{\sqrt{3+13}+4} {\sqrt{3+6}+3} =\frac{\sqrt{16}+4} {\sqrt 9 + 3} =\frac 8 6 =\frac 4 3$$.

$$\displaystyle \lim_{x\to 3}\, \frac{\sqrt{x+6} - 3} {\sqrt{x+13} - 4} =\frac 4 3$$