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Limits and Their Properties

1.3 Evaluating Limits Analytically

Focus on direct substitution and algebraic simplification, trigonometric limits, the squeeze theorem.

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Ideas in this lesson

01Direct substitution and algebraic simplificationWorked steps · five questions02Trigonometric limitsWorked steps · five questions03The Squeeze TheoremWorked steps · five questions
Idea 1 of 3

Direct substitution and algebraic simplification

Continuous rules allow direct substitution. A00A \frac{0}{0} form requires analysis; it is not a limit value.

Worked example 1

Evaluate lim⁡x→2(x2+1)\displaystyle \lim\limits_{x\to 2} (x^{2} + 1).
  1. f(2)=22+1f(2) = 2^{2} + 1
  2. lim⁡x→2(x2+1)=5\displaystyle \lim\limits_{x\to 2} (x^{2} + 1)= 5

Answer: 55

Worked example 2

Evaluate lim⁡x→3x2−9x−3\displaystyle \lim\limits_{x\to 3} \dfrac{x^{2} - 9}{x - 3}.
  1. x2−9=(x−3)(x+3)x^{2} - 9 = (x - 3)(x + 3)
  2. x2−9x−3=x+3(x≠3)\dfrac{x^{2} - 9}{x - 3} = x + 3 (x \ne 3)
  3. lim⁡x→3(x+3)=6\displaystyle \lim\limits_{x\to 3} (x + 3)= 6

Answer: 66

Idea 2 of 3

Trigonometric limits

In radians, sin⁡uu\frac{\sin u}{u} tends to 11 near zero. Adjust the inner scale before applying this limit.

Worked example 1

Evaluate lim⁡x→0sin⁡(2x)x(\displaystyle \lim\limits_{x\to 0} \dfrac{\sin (2x)}{x} ( angles in radians )).
  1. sin⁡(2x)x=2sin⁡(2x)2x\dfrac{\sin (2x)}{x} = 2\dfrac{\sin (2x)}{2x}
  2. 2⋅1=22 \cdot 1 = 2

Answer: 22

Worked example 2

Evaluate lim⁡x→01−cos⁡(3x)x2(\displaystyle \lim\limits_{x\to 0} \dfrac{1 - \cos (3x)}{x^{2}} ( angles in radians )).
  1. 1−cos⁡(3x)x2=sin⁡2(3x)x2(1+cos⁡(3x))\dfrac{1 - \cos (3x)}{x^{2}} = \dfrac{\sin ^{2}(3x)}{x^{2}(1 + \cos (3x))}
  2. 91+1=92≈4.5\dfrac{9}{1 + 1} = \dfrac{9}{2} \approx 4.5

Answer: 92≈4.5\dfrac{9}{2} \approx 4.5

Idea 3 of 3

The Squeeze Theorem

If two bounding functions have the same limit, any function trapped between them has that limit.

Worked example 1

Use the Squeeze Theorem to evaluate lim⁡x→0xsin⁡(2x)\displaystyle \lim\limits_{x\to 0} x \sin (\dfrac{2}{x}).
  1. −∣x∣≤xsin⁡(2x)≤∣x∣-|x| \le x \sin (\dfrac{2}{x}) \le |x|
  2. lim⁡x→0∣x∣=0\displaystyle \lim\limits_{x\to 0} |x| = 0
  3. lim⁡x→0xsin⁡(2x)=0\displaystyle \lim\limits_{x\to 0} x \sin (\dfrac{2}{x}) = 0

Answer: 00

Worked example 2

Use the Squeeze Theorem to evaluate lim⁡x→0x2sin⁡(3x)\displaystyle \lim\limits_{x\to 0} x^{2} \sin (\dfrac{3}{x}).
  1. −∣x∣2≤x2sin⁡(3x)≤∣x∣2-|x|^{2} \le x^{2} \sin (\dfrac{3}{x}) \le |x|^{2}
  2. lim⁡x→0∣x∣2=0\displaystyle \lim\limits_{x\to 0} |x|^{2} = 0
  3. lim⁡x→0x2sin⁡(3x)=0\displaystyle \lim\limits_{x\to 0} x^{2} \sin (\dfrac{3}{x}) = 0

Answer: 00

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §1.3,p1.3, p. 5959. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.