Compute limits with algebra

About this lesson

Precalculus › Limits: an introduction

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Compute limits with algebra

Warm-up

Three quick questions from earlier lessons.

Discover

To find lim⁡x→1(x2+3x)\displaystyle\lim_{x \to 1} (x^2 + 3x), what can you do?

  1. Substitute x=1x = 1: the limit is 4
  2. You can never substitute in a limit

Key idea

First try substituting. If you get 00\dfrac{0}{0}, simplify and try again.

If substituting gives a number, that is the limit (for polynomials, and rational functions where the denominator is not 0). If it gives 00\dfrac{0}{0}: factor and cancel, or expand, or use (x−s)(x+s)=x−s2(\sqrt{x} - s)(\sqrt{x} + s) = x - s^2; then substitute.

Worked example

Find lim⁡h→0(3+h)2−9h\displaystyle\lim_{h \to 0} \dfrac{(3 + h)^2 - 9}{h}.

(3+h)2−9=6h+h2(3 + h)^2 - 9 = 6h + h^2
Substitution gives 0/0: expand instead.
6h+h2h=6+h\dfrac{6h + h^2}{h} = 6 + h
Divide by h (h ≠ 0).
The limit is 66
h → 0.

Your turn, guided

Your turn: lim⁡x→4x2−5x+4x−4\displaystyle\lim_{x \to 4} \dfrac{x^2 - 5x + 4}{x - 4}.

=lim⁡x→4(x−1)= \displaystyle\lim_{x \to 4} (x - 1)
Factor: (x − 4)(x − 1).
=3= 3
4 − 1.

Practice ★

  1. Find lim⁡x→7x2−15x+56x−7\displaystyle\lim_{x \to 7} \dfrac{x^2 - 15x + 56}{x - 7}.
  2. Find lim⁡x→1x2−4x+3x−1\displaystyle\lim_{x \to 1} \dfrac{x^2 - 4x + 3}{x - 1}.
  3. Find lim⁡x→2x2−9x+14x−2\displaystyle\lim_{x \to 2} \dfrac{x^2 - 9x + 14}{x - 2}.
  4. Find lim⁡x→4x2−11x+28x−4\displaystyle\lim_{x \to 4} \dfrac{x^2 - 11x + 28}{x - 4}.

Watch out!

Practice ★★

  1. Find lim⁡h→0(8+h)2−64h\displaystyle\lim_{h \to 0} \dfrac{(8 + h)^2 - 64}{h}.
  2. Find lim⁡h→0(7+h)2−49h\displaystyle\lim_{h \to 0} \dfrac{(7 + h)^2 - 49}{h}.
  3. Find lim⁡h→0(4+h)2−16h\displaystyle\lim_{h \to 0} \dfrac{(4 + h)^2 - 16}{h}.
  1. Find lim⁡x→9x−3x−9\displaystyle\lim_{x \to 9} \dfrac{\sqrt{x} - 3}{x - 9}.
  2. Find lim⁡x→4x−2x−4\displaystyle\lim_{x \to 4} \dfrac{\sqrt{x} - 2}{x - 4}.

Exit ticket

Three questions to check you have got it.

  1. Find lim⁡x→7x2−15x+56x−7\displaystyle\lim_{x \to 7} \dfrac{x^2 - 15x + 56}{x - 7}.
  2. Find lim⁡x→1x2−4x+3x−1\displaystyle\lim_{x \to 1} \dfrac{x^2 - 4x + 3}{x - 1}.
  3. Find lim⁡x→2x2−9x+14x−2\displaystyle\lim_{x \to 2} \dfrac{x^2 - 9x + 14}{x - 2}.

Summary card

Limits with algebra

  1. Substitute. A number? That is the limit.
  2. 00\dfrac{0}{0}? Factor and cancel, expand, or use conjugates.
  3. Substitute again.

Watch out: 00\dfrac{0}{0} is never the answer.

End of the lesson

Next lesson: One-sided limits Back to the chapter

Answers: Compute limits with algebra

A

  1. −1-1
  2. −2-2
  3. −5-5
  4. −3-3

B

  1. 1616
  2. 1414
  3. 88

C

  1. 16\frac{1}{6}
  2. 14\frac{1}{4}

D

  1. −1-1
  2. −2-2
  3. −5-5

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