Dove's LibraryDove's Library

1.7 - Selecting Procedures for Determining Limits

Overview

We have a bunch of algebraic tools for evaluating limits. The trick is picking the right one for each problem. Here we add two more powerful techniques:
  1. Rationalizing (multiplying by conjugates)
  1. Complex fractions (fractions within fractions)
General strategy: always try direct substitution first. If it produces an indeterminate form like 00\frac{0}{0}, then pick an algebraic manipulation technique.

Review: Direct Substitution

Example 1: find lim⁡x→3(x2+4x−7)\lim_{x \to 3} (x^2 + 4x - 7).
lim⁡x→3(x2+4x−7)=(3)2+4(3)−7=9+12−7=14\lim_{x \to 3} (x^2 + 4x - 7) = (3)^2 + 4(3) - 7 = 9 + 12 - 7 = 14
Answer: 1414. Direct substitution worked because we got a defined value.

Method 3: Rationalizing (Multiplying by the Conjugate)

When a limit involves square roots and produces 00\frac{0}{0}, we can rationalize by multiplying by the conjugate.
Key identity (difference of squares):
(a−b)(a+b)=a2−b2(a - b)(a + b) = a^2 - b^2
Two expressions are conjugates if they have the form a−ba - b and a+ba + b. One has addition, the other subtraction, and when you multiply them the middle terms cancel.
Examples of conjugate pairs: - x−5\sqrt{x} - 5 and x+5\sqrt{x} + 5 - x+3+2\sqrt{x + 3} + 2 and x+3−2\sqrt{x + 3} - 2 - 4−x−14 - \sqrt{x - 1} and 4+x−14 + \sqrt{x - 1}
Example 2: radical in the numerator
Find lim⁡x→7x+2−3x−7\lim_{x \to 7} \frac{\sqrt{x + 2} - 3}{x - 7}.
Step 1, try direct substitution:
7+2−37−7=9−30=3−30=00\frac{\sqrt{7 + 2} - 3}{7 - 7} = \frac{\sqrt{9} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0}
Indeterminate, so we manipulate.
Step 2, identify the conjugate. The numerator is x+2−3\sqrt{x + 2} - 3, so the conjugate is x+2+3\sqrt{x + 2} + 3 (just flip the sign).
Step 3, multiply by conjugateconjugate\frac{\text{conjugate}}{\text{conjugate}}:
lim⁡x→7x+2−3x−7⋅x+2+3x+2+3\lim_{x \to 7} \frac{\sqrt{x + 2} - 3}{x - 7} \cdot \frac{\sqrt{x + 2} + 3}{\sqrt{x + 2} + 3}
Step 4, apply difference of squares to the numerator:
=lim⁡x→7(x+2)2−(3)2(x−7)(x+2+3)=lim⁡x→7(x+2)−9(x−7)(x+2+3)= \lim_{x \to 7} \frac{(\sqrt{x + 2})^2 - (3)^2}{(x - 7)(\sqrt{x + 2} + 3)} = \lim_{x \to 7} \frac{(x + 2) - 9}{(x - 7)(\sqrt{x + 2} + 3)}=lim⁡x→7x−7(x−7)(x+2+3)= \lim_{x \to 7} \frac{x - 7}{(x - 7)(\sqrt{x + 2} + 3)}
Step 5, cancel:
=lim⁡x→71x+2+3= \lim_{x \to 7} \frac{1}{\sqrt{x + 2} + 3}
Step 6, substitute:
=17+2+3=13+3=16= \frac{1}{\sqrt{7 + 2} + 3} = \frac{1}{3 + 3} = \frac{1}{6}
Answer: 16\frac{1}{6}
Example 3: radical in the denominator
Find lim⁡x→5x−5x+4−3\lim_{x \to 5} \frac{x - 5}{\sqrt{x + 4} - 3}.
Step 1, check the form:
5−55+4−3=09−3=00\frac{5 - 5}{\sqrt{5 + 4} - 3} = \frac{0}{\sqrt{9} - 3} = \frac{0}{0}
Step 2, multiply by the conjugate of the denominator, x+4+3\sqrt{x + 4} + 3:
lim⁡x→5x−5x+4−3⋅x+4+3x+4+3\lim_{x \to 5} \frac{x - 5}{\sqrt{x + 4} - 3} \cdot \frac{\sqrt{x + 4} + 3}{\sqrt{x + 4} + 3}
Step 3, difference of squares in the denominator:
=lim⁡x→5(x−5)(x+4+3)(x+4)−9=lim⁡x→5(x−5)(x+4+3)x−5= \lim_{x \to 5} \frac{(x - 5)(\sqrt{x + 4} + 3)}{(x + 4) - 9} = \lim_{x \to 5} \frac{(x - 5)(\sqrt{x + 4} + 3)}{x - 5}
Step 4, cancel (x−5)(x - 5):
=lim⁡x→5(x+4+3)= \lim_{x \to 5} (\sqrt{x + 4} + 3)
Step 5, substitute:
=9+3=3+3=6= \sqrt{9} + 3 = 3 + 3 = 6
Answer: 66
Example 4: careful with negative signs
Find lim⁡x→12x−124−x+4\lim_{x \to 12} \frac{x - 12}{4 - \sqrt{x + 4}}.
Step 1, verify the form:
12−124−16=04−4=00\frac{12 - 12}{4 - \sqrt{16}} = \frac{0}{4 - 4} = \frac{0}{0}
Step 2, the conjugate of 4−x+44 - \sqrt{x + 4} is 4+x+44 + \sqrt{x + 4}:
lim⁡x→12x−124−x+4⋅4+x+44+x+4\lim_{x \to 12} \frac{x - 12}{4 - \sqrt{x + 4}} \cdot \frac{4 + \sqrt{x + 4}}{4 + \sqrt{x + 4}}
Step 3, difference of squares:
=lim⁡x→12(x−12)(4+x+4)(4)2−(x+4)2=lim⁡x→12(x−12)(4+x+4)16−(x+4)= \lim_{x \to 12} \frac{(x - 12)(4 + \sqrt{x + 4})}{(4)^2 - (\sqrt{x + 4})^2} = \lim_{x \to 12} \frac{(x - 12)(4 + \sqrt{x + 4})}{16 - (x + 4)}
Step 4, simplify the denominator:
=lim⁡x→12(x−12)(4+x+4)12−x= \lim_{x \to 12} \frac{(x - 12)(4 + \sqrt{x + 4})}{12 - x}
Step 5, factor out −1-1 from the denominator, since 12−x=−(x−12)12 - x = -(x - 12):
=lim⁡x→12(x−12)(4+x+4)−(x−12)= \lim_{x \to 12} \frac{(x - 12)(4 + \sqrt{x + 4})}{-(x - 12)}
Step 6, cancel and simplify:
=lim⁡x→12−(4+x+4)= \lim_{x \to 12} -(4 + \sqrt{x + 4})
Step 7, substitute:
=−(4+16)=−(4+4)=−8= -(4 + \sqrt{16}) = -(4 + 4) = -8
Answer: −8-8

Method 4: Complex Fractions

A complex fraction is a fraction with one or more fractions inside its numerator or denominator.
abcd\frac{\frac{a}{b}}{\frac{c}{d}}
Strategy: 1. Find common denominators inside the numerator and denominator separately 2. Combine the fractions 3. Remember that dividing by a fraction is the same as multiplying by its reciprocal
Example 5: basic complex fraction
Find lim⁡x→0x13−1x+3\lim_{x \to 0} \frac{x}{\frac{1}{3} - \frac{1}{x + 3}}.
Step 1, check the form:
013−13=00\frac{0}{\frac{1}{3} - \frac{1}{3}} = \frac{0}{0}
Step 2, get a common denominator in the bottom. The complex part is 13−1x+3\frac{1}{3} - \frac{1}{x + 3}, and the common denominator is 3(x+3)3(x + 3):
13=x+33(x+3)1x+3=33(x+3)\frac{1}{3} = \frac{x + 3}{3(x + 3)} \qquad \frac{1}{x + 3} = \frac{3}{3(x + 3)}
Step 3, combine:
13−1x+3=(x+3)−33(x+3)=x3(x+3)\frac{1}{3} - \frac{1}{x + 3} = \frac{(x + 3) - 3}{3(x + 3)} = \frac{x}{3(x + 3)}
Step 4, rewrite the limit:
lim⁡x→0xx3(x+3)\lim_{x \to 0} \frac{x}{\frac{x}{3(x + 3)}}
Step 5, division becomes multiplication by the reciprocal:
=lim⁡x→0x⋅3(x+3)x= \lim_{x \to 0} x \cdot \frac{3(x + 3)}{x}
Step 6, cancel xx:
=lim⁡x→03(x+3)= \lim_{x \to 0} 3(x + 3)
Step 7, substitute:
=3(0+3)=9= 3(0 + 3) = 9
Answer: 99
Example 6: a harder complex fraction
Find lim⁡x→019−1(x+3)2x\lim_{x \to 0} \frac{\frac{1}{9} - \frac{1}{(x + 3)^2}}{x}.
Step 1, verify the form. At x=0x = 0: 19−190=00\frac{\frac{1}{9} - \frac{1}{9}}{0} = \frac{0}{0}
Step 2, common denominator for the numerator is 9(x+3)29(x + 3)^2:
19=(x+3)29(x+3)21(x+3)2=99(x+3)2\frac{1}{9} = \frac{(x + 3)^2}{9(x + 3)^2} \qquad \frac{1}{(x + 3)^2} = \frac{9}{9(x + 3)^2}
Step 3, combine:
(x+3)2−99(x+3)2\frac{(x + 3)^2 - 9}{9(x + 3)^2}
Step 4, expand (x+3)2=x2+6x+9(x + 3)^2 = x^2 + 6x + 9, so the numerator becomes:
(x2+6x+9)−99(x+3)2=x2+6x9(x+3)2\frac{(x^2 + 6x + 9) - 9}{9(x + 3)^2} = \frac{x^2 + 6x}{9(x + 3)^2}
Step 5, rewrite the original limit:
lim⁡x→0x2+6x9(x+3)2x\lim_{x \to 0} \frac{\frac{x^2 + 6x}{9(x + 3)^2}}{x}
Step 6, multiply by the reciprocal:
=lim⁡x→0x2+6x9(x+3)2⋅1x=lim⁡x→0x2+6x9x(x+3)2= \lim_{x \to 0} \frac{x^2 + 6x}{9(x + 3)^2} \cdot \frac{1}{x} = \lim_{x \to 0} \frac{x^2 + 6x}{9x(x + 3)^2}
Step 7, factor the numerator:
=lim⁡x→0x(x+6)9x(x+3)2= \lim_{x \to 0} \frac{x(x + 6)}{9x(x + 3)^2}
Step 8, cancel xx:
=lim⁡x→0x+69(x+3)2= \lim_{x \to 0} \frac{x + 6}{9(x + 3)^2}
Step 9, substitute:
=0+69(0+3)2=69(9)=681= \frac{0 + 6}{9(0 + 3)^2} = \frac{6}{9(9)} = \frac{6}{81}
Step 10, simplify:
=227= \frac{2}{27}
Answer: 227\frac{2}{27}

Strategy for Selecting Procedures

Step 1: try direct substitution - If it works, you’re done - If it gives 00\frac{0}{0}, go to Step 2
Step 2: identify the structure - Square roots → try rationalizing (multiply by the conjugate) - Fractions within fractions → simplify the complex fraction (common denominators) - Factorable polynomials → factor and cancel - Trig functions as x→0x \to 0 → use the special trig limits
Other situations to recognize - Trig functions that don’t match the special limits → rewrite in sine and cosine first - Piecewise function at a boundary → find both one-sided limits using the matching pieces - Absolute value → look at each side separately with one-sided limits - A bounded, oscillating factor (like sin⁡(1x)\sin\left(\frac{1}{x}\right)) times something that goes to 00 → Squeeze Theorem - nonzero0\frac{\text{nonzero}}{0} → no algebra will cancel it; check the sign on each side to see whether it goes to ∞\infty or −∞-\infty (if the sides disagree, the limit does not exist)

Practice Problems

Problem 1: find lim⁡x→8x+1−3x−8\lim_{x \to 8} \frac{\sqrt{x + 1} - 3}{x - 8}
Conjugate: x+1+3\sqrt{x + 1} + 3
(x+1)−9(x−8)(x+1+3)=x−8(x−8)(x+1+3)\frac{(x + 1) - 9}{(x - 8)(\sqrt{x + 1} + 3)} = \frac{x - 8}{(x - 8)(\sqrt{x + 1} + 3)}=18+1+3=13+3=16= \frac{1}{\sqrt{8 + 1} + 3} = \frac{1}{3 + 3} = \frac{1}{6}
Problem 2: find lim⁡x→0x14−1x+4\lim_{x \to 0} \frac{x}{\frac{1}{4} - \frac{1}{x + 4}}
Common denominator: 4(x+4)4(x + 4)
14−1x+4=x+4−44(x+4)=x4(x+4)\frac{1}{4} - \frac{1}{x + 4} = \frac{x + 4 - 4}{4(x + 4)} = \frac{x}{4(x + 4)}xx4(x+4)=x⋅4(x+4)x=4(x+4)\frac{x}{\frac{x}{4(x + 4)}} = x \cdot \frac{4(x + 4)}{x} = 4(x + 4)
At x=0x = 0: 4(4)=164(4) = 16
Problem 3: find lim⁡x→3x−32−x+1\lim_{x \to 3} \frac{x - 3}{2 - \sqrt{x + 1}}
Conjugate: 2+x+12 + \sqrt{x + 1}
Denominator becomes 4−(x+1)=3−x=−(x−3)4 - (x + 1) = 3 - x = -(x - 3)
(x−3)(2+x+1)−(x−3)=−(2+x+1)\frac{(x - 3)(2 + \sqrt{x + 1})}{-(x - 3)} = -(2 + \sqrt{x + 1})
At x=3x = 3: −(2+2)=−4-(2 + 2) = -4

Key Takeaways

  • Always try direct substitution first, it’s the fastest method
  • Conjugates eliminate radicals through the difference of squares formula
  • When the radical is in the numerator, multiply by its conjugate
  • When the radical is in the denominator, multiply by its conjugate
  • Watch the negative signs when simplifying, and factor out −1-1 when needed
  • Complex fractions need common denominators first
  • After combining fractions, multiply by the reciprocal instead of dividing
  • Factor whenever possible to reveal common terms that cancel
  • Be precise with your algebra, these problems have a lot of steps
  • After all the manipulation, always finish with direct substitution

AP® is a registered trademark of the College Board, which is not affiliated with Dove's Library and was not involved in the production of, and does not endorse, this website.

Providing free, high-quality AP notes to help students succeed, earn college credit, and reach their educational goals, ensuring everyone has access to the resources they need.

© 2026 Standard Learning Group, Inc., which operates Dove's Library. All rights reserved.