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Applications of Differentiation

3.5 Limits at Infinity

Focus on slant asymptotes, leading terms at infinity, horizontal asymptotes, different numerator and denominator degrees.

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Ideas in this lesson

01Slant asymptotesWorked steps · five questions02Leading terms at infinityWorked steps · five questions03Horizontal asymptotesWorked steps · five questions04Different numerator and denominator degreesWorked steps · five questions
Idea 1 of 4

Slant asymptotes

Polynomial division gives f(x)=mx+c+af(x) = mx + c + a remainder tending to zero. The slant asymptote is y=mx+cy = mx + c.

Worked example 1

For f(x)=x2+2x+1xf(x) = \dfrac{x^{2} + 2x + 1}{x}, find the yy-intercept of its slant asymptote.
  1. f(x)=x+2+1xf(x) = x + 2 + \dfrac{1}{x}
  2. lim⁡x→±∞[f(x)−(x+2)]=lim⁡1x=0\displaystyle \lim\limits_{x\to \pm \infty } [f(x) - (x + 2)] = \lim \dfrac{1}{x} = 0
  3. Slant asymptote: y=x+2y = x + 2; yy-intercept =2= 2

Answer: 22

Worked example 2

For f(x)=x2+3x+2xf(x) = \dfrac{x^{2} + 3x + 2}{x}, find the yy-intercept of its slant asymptote.
  1. f(x)=x+3+2xf(x) = x + 3 + \dfrac{2}{x}
  2. lim⁡x→±∞[f(x)−(x+3)]=lim⁡2x=0\displaystyle \lim\limits_{x\to \pm \infty } [f(x) - (x + 3)] = \lim \dfrac{2}{x} = 0
  3. Slant asymptote: y=x+3y = x + 3; yy-intercept =3= 3

Answer: 33

Idea 2 of 4

Leading terms at infinity

Divide a rational expression by its highest relevant power. Compare the surviving terms rather than substituting infinity.

Worked example 1

Evaluate lim⁡x→∞2x2+12x2+3\displaystyle \lim\limits_{x\to \infty } \dfrac{2x^{2} + 1}{2x^{2} + 3}.
  1. 2x2+12x2+3=2+1x22+3x2\dfrac{2x^{2} + 1}{2x^{2} + 3} = \dfrac{2 + \dfrac{1}{x^{2}}}{2 + \dfrac{3}{x^{2}}}
  2. 1x→0,1x2→0(x→∞)\dfrac{1}{x} \to 0, \dfrac{1}{x^{2}} \to 0 (x \to \infty )
  3. 2+02+0=1\dfrac{2 + 0}{2 + 0} = 1

Answer: 11

Worked example 2

Evaluate lim⁡x→∞x+3x2+1\displaystyle \lim\limits_{x\to \infty } \dfrac{x + 3}{x^{2} + 1}.
  1. x+3x2+1=1x+3x21+1x2\dfrac{x + 3}{x^{2} + 1} = \dfrac{\dfrac{1}{x} + \dfrac{3}{x^{2}}}{1 + \dfrac{1}{x^{2}}}
  2. 1x→0,1x2→0(x→∞)\dfrac{1}{x} \to 0, \dfrac{1}{x^{2}} \to 0 (x \to \infty )
  3. 0+01+0=0\dfrac{0 + 0}{1 + 0} = 0

Answer: 00

Idea 3 of 4

Horizontal asymptotes

A finite limit at positive or negative infinity defines a horizontal asymptote. A graph may cross it.

Worked example 1

Find the horizontal asymptote height of f(x)=2x2−1x2+2f(x) = \dfrac{2x^{2} - 1}{x^{2} + 2}.
  1. f(x)=2−1x21+2x2f(x) = \dfrac{2 - \dfrac{1}{x^{2}}}{1 + \dfrac{2}{x^{2}}}
  2. lim⁡x→±∞f(x)=21\displaystyle \lim\limits_{x\to \pm \infty } f(x) = \dfrac{2}{1}
  3. Horizontal asymptote: y=2y = 2

Answer: 22

Worked example 2

Find the horizontal asymptote height of f(x)=3x2−12x2+2f(x) = \dfrac{3x^{2} - 1}{2x^{2} + 2}.
  1. f(x)=3−1x22+2x2f(x) = \dfrac{3 - \dfrac{1}{x^{2}}}{2 + \dfrac{2}{x^{2}}}
  2. lim⁡x→±∞f(x)=32\displaystyle \lim\limits_{x\to \pm \infty } f(x) = \dfrac{3}{2}
  3. Horizontal asymptote: y=32≈1.5y = \dfrac{3}{2} \approx 1.5

Answer: 32≈1.5\dfrac{3}{2} \approx 1.5

Idea 4 of 4

Different numerator and denominator degrees

A lower numerator degree gives limit zero; equal degrees give a coefficient ratio; a higher degree requires further analysis.

Worked example 1

Evaluate lim⁡x→∞2x2+12x2+3\displaystyle \lim\limits_{x\to \infty } \dfrac{2x^{2} + 1}{2x^{2} + 3}.
  1. 2x2+12x2+3=2+1x22+3x2\dfrac{2x^{2} + 1}{2x^{2} + 3} = \dfrac{2 + \dfrac{1}{x^{2}}}{2 + \dfrac{3}{x^{2}}}
  2. 1x→0,1x2→0(x→∞)\dfrac{1}{x} \to 0, \dfrac{1}{x^{2}} \to 0 (x \to \infty )
  3. 2+02+0=1\dfrac{2 + 0}{2 + 0} = 1

Answer: 11

Worked example 2

Evaluate lim⁡x→∞x+3x2+1\displaystyle \lim\limits_{x\to \infty } \dfrac{x + 3}{x^{2} + 1}.
  1. x+3x2+1=1x+3x21+1x2\dfrac{x + 3}{x^{2} + 1} = \dfrac{\dfrac{1}{x} + \dfrac{3}{x^{2}}}{1 + \dfrac{1}{x^{2}}}
  2. 1x→0,1x2→0(x→∞)\dfrac{1}{x} \to 0, \dfrac{1}{x^{2}} \to 0 (x \to \infty )
  3. 0+01+0=0\dfrac{0 + 0}{1 + 0} = 0

Answer: 00

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §3.5,p3.5, p. 195195. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.