Precalculus by Richard Wright

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Then Jesus said to his disciples, “Whoever wants to be my disciple must deny themselves and take up their cross and follow me. For whoever wants to save their life will lose it, but whoever loses their life for me will find it.” Matthew‬ ‭16‬:‭24‬-‭25‬ ‭NIV‬‬

12-04 Limits at Infinity and Limits of Sequences

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.4.2

soccer ball
Figure 1: credit (Pixnio/Bicanski)

Derivatives, or rates-of-change, are the first major part of calculus. Finding integrals, or areas, are the second major part. Calculating an integral involves adding together an infinite number of tiny areas. For example, the surface area of the soccer ball can be estimated by adding up the area of the colored polygons. A better estimate could be found by adding up the areas of the smaller leather polygons that are stitched together. An even better estimate would be to split the area into an infinite number of infinitely tiny areas, and add those areas together. This is done by taking the limit of a function as x approaches infinity. That limit is the focus of this lesson.

Limits at Infinity

A limit at infinity can be found by using the fact that \(f(x) = \frac{1}{x}\) has a horizontal asymptote at y = 0. That means that as x approaches infinity, the function approaches zero. The same is true if f(x) has any exponent. Some limits at infinity can then be evaluating by splitting them into several simplified fractions where \(x^n\) is the denominator of each fraction.

$$ \lim_{x \rightarrow ∞} \frac{1}{x^n} = 0 $$

$$ \lim_{x \rightarrow -∞} \frac{1}{x^n} = 0 $$

Example 1: Evaluate a Limit at Infinity (Long Method)

Evaluate \(\displaystyle \lim_{x \rightarrow ∞} \frac{1 - 3x + 2x^2}{x^2}\).

Solution

Start by splitting the function into separate fractions with the same denominator.

$$ \lim_{x \rightarrow ∞} \frac{1 - 3x + 2x^2}{x^2} $$

$$ \lim_{x \rightarrow ∞} \left(\frac{1}{x^2} - \frac{3x}{x^2} + \frac{2x^2}{x^2}\right) $$

Simplify each fraction.

$$ \lim_{x \rightarrow ∞} \left(\frac{1}{x^2} - \frac{3}{x} + 2\right) $$

Evaluate the limit of each term.

$$ = 0 - 0 + 2 $$

$$ = 2 $$

Try It 1

Evaluate \(\displaystyle \lim_{x \rightarrow ∞} \frac{5x^3 - 2x^2 + 3x - 14}{x^3}\).

Answer

5

The limit at infinity is the same as the value of the horizontal asymptote for the function. Thus, the same shortcuts for finding horizontal asymptotes also work to find limits at infinity.

Shortcuts for Evaluating Limits at Infinity

If f(x) is a rational function, then \(\displaystyle \lim_{x \rightarrow ∞} f(x)\) is found by

Let N = degree of the numerator

Let D = degree of the denominator

Example 2: Evaluate a Limit at Infinity (Shortcut)

Evaluate

  1. \(\displaystyle \lim_{x \rightarrow ∞} \frac{x - 3}{2x^2 + 4x + 5}\)
  2. \(\displaystyle \lim_{x \rightarrow ∞} \frac{x^2 - 3}{2x^2 + 4x + 5}\)
  3. \(\displaystyle \lim_{x \rightarrow ∞} \frac{x^3 - 3}{2x^2 + 4x + 5}\)
Solution
  1. Start by finding the degree of the numerator (N) and the degree of the denominator (D).

    $$ \lim_{x \rightarrow ∞} \frac{x^{\color{blue}{1}} - 3}{2x^{\color{red}{2}} + 4x + 5} $$

    N = 1; D = 2

    N < D, so \(\displaystyle \lim_{x \rightarrow ∞} \frac{x - 3}{2x^2 + 4x + 5} = 0\).

    graph
    Figure 2: The horizontal asymptote is y = 0 and limit at infinity is 0.
  2. Start by finding the degree of the numerator (N) and the degree of the denominator (D).

    $$ \lim_{x \rightarrow ∞} \frac{x^2 - 3}{2x^2 + 4x + 5} $$

    N = 2; D = 2

    N = D, so the limit is the leading coefficients. \(\displaystyle \lim_{x \rightarrow ∞} \frac{x^2 - 3}{2x^2 + 4x + 5} = \frac{1}{2}\).

    graph
    Figure 3: The horizontal asymptote is y = \(\frac{1}{2}\) and limit at infinity is \(\frac{1}{2}\).
  3. Start by finding the degree of the numerator (N) and the degree of the denominator (D).

    $$ \lim_{x \rightarrow ∞} \frac{x^3 - 3}{2x^2 + 4x + 5} $$

    N = 3; D = 2

    N > D, so the limit does not exist. \(\displaystyle \lim_{x \rightarrow ∞} \frac{x^3 - 3}{2x^2 + 4x + 5}\) does not exist.

    graph
    Figure 4: There is no horizontal asymptote and limit at infinity does not exist.
Try It 2

Evaluate \(\displaystyle \lim_{x \rightarrow ∞} \frac{3x^4 + 2x^3 - 5x + 10}{2x^4 + 16}\)

Answer

\(\frac{3}{2}\)

Limits of Sequences

The limit of a sequence is the value the terms tend to approach. If the sequences terms approach a single number, then the sequence converges. If the sequence terms do not approach a single number, then the sequence diverges. If a sequence converges, then the limit of the sequence is found by

$$ \lim_{n \rightarrow ∞} a_n $$

graph
Figure 5: The terms of the sequence tend towards 0.
Limit of a Sequence

$$ \lim_{n \rightarrow ∞} a_n $$

If the limit exists, then the sequence converges.

If the limit does not exist, then the sequence diverges.

Example 3: Find the Limit of a Sequence

Find the limit of the sequence \(a_n = \frac{(n + 2)(6n - 5)}{2 - 3n + 3n^2}\) and state whether it converges or diverges.

Solution

Find the limit of the sequence as n approaches infinity.

$$ \lim_{n \rightarrow ∞} \frac{(n + 2)(6n - 5)}{2 - 3n + 3n^2} $$

Use the shortcut method to find the limit. Multiply the numerator to find its degree and rearrange the denominator with decreasing exponents.

$$ \lim_{n \rightarrow ∞} \frac{6n^2 + 7n - 10}{3n^2 - 3n + 2} $$

N = 2; D = 2

N = D, so the limit is the leading coefficients. \(\displaystyle \lim_{n \rightarrow ∞} \frac{6n^2 + 7n - 10}{3n^2 - 3n + 2} = \frac{6}{3} = 2\).

Since the limit of the sequence exists, it converges to 2

graph
Figure 6: The horizontal asymptote is an = 2 and limit at infinity is 2.

Example 4: Find the Limit of a Sequence

Find the limit of the sequence \(a_n = \frac{4}{n^3} \cdot \left[\frac{n(n+1)(2n+1)}{6}\right]\) and state whether it converges or diverges.

Solution

Find the limit of the sequence as n approaches infinity.

$$ \lim_{n \rightarrow ∞} \frac{4}{n^3} \cdot \left[\frac{n(n+1)(2n+1)}{6}\right] $$

Use the shortcut method to find the limit. Multiply the numerator and denominator to find their degrees.

$$ \lim_{n \rightarrow ∞} \frac{4}{n^3} \cdot \left[\frac{2n^3 + 3n^2 + n}{6}\right] $$

$$ \lim_{n \rightarrow ∞} \frac{8n^3 + 12n^2 + 4n}{6n^3} $$

N = 3; D = 3

N = D, so the limit is the leading coefficients. \(\displaystyle \lim_{n \rightarrow ∞} \frac{8n^3 + 12n^2 + 4n}{6n^3} = \frac{8}{6} = \frac{4}{3}\).

Since the limit of the sequence exists, it converges to \(\frac{4}{3}\).

graph
Figure 7: The horizontal asymptote is an = \(\frac{4}{3}\) and limit at infinity is \(\frac{4}{3}\).
Try It 3

Find the limit of the sequence \(a_n = \frac{3}{n^2} \cdot \left[\frac{n(n+1)}{2}\right]\) and state whether it converges or diverges.

Answer

The sequences converges to \(\frac{3}{2}\).

Lesson Summary

Shortcuts for Evaluating Limits at Infinity

If f(x) is a rational function, then \(\displaystyle \lim_{x \rightarrow ∞} f(x)\) is found by

Let N = degree of the numerator

Let D = degree of the denominator


Limit of a Sequence

$$ \lim_{n \rightarrow ∞} a_n $$

If the limit exists, then the sequence converges.

If the limit does not exist, then the sequence diverges.

Helpful videos about this lesson.

Practice Exercises

  1. What does it mean to find the limit of a sequence?
  2. Evaluate the limit if it exists.

  3. \(\displaystyle \lim_{x \rightarrow ∞} \frac{6 - 2x + x^2 - 4x^3}{x^3}\)
  4. \(\displaystyle \lim_{x \rightarrow ∞} \frac{x^4 + 2x^2 - 1}{2 + 5x + 3x^3}\)
  5. \(\displaystyle \lim_{x \rightarrow ∞} \frac{17 - 3x^3 + x^2}{6x^3}\)
  6. \(\displaystyle \lim_{x \rightarrow ∞} \frac{(x + 4)(x - 9)}{(2x + 1)(3x - 4)}\)
  7. Find the limit of the sequence and state whether it converges.

  8. \(a_n = \frac{(3n - 1)(n^2 + 4)}{2 - 4n - 5n^2}\)
  9. \(a_n = \frac{(n^2 + 2n - 1)(3n + 2)}{5 - 2n - 9n^3}\)
  10. \(a_n = \frac{10}{n^4} \cdot \left[\frac{n^2(n+1)^2}{4}\right]\)
  11. \(a_n = \frac{20}{n^3} \cdot \left[\frac{n(n+1)(2n+1)}{6}\right]\)
  12. \(a_n = \frac{5}{n^2} \cdot \left[\frac{n(n+1)}{2}\right]\)
  13. Mixed Review

  14. (12-03) Find the derivative of \(f(x) = 2x^3 - 3x\).
  15. (12-03) Find the derivative of \(f(x) = 4\sqrt{x} + 5\).
  16. (12-02) Evaluate \(\displaystyle \lim_{x \rightarrow -3} \frac{x + 3}{x^2 + 2x - 3}\).
  17. (12-01) Evaluate \(\displaystyle \lim_{x \rightarrow 2} \frac{x + 3}{x^2 + 2x - 3}\).
  18. (10-02) Evaluate \(\displaystyle \sum_{n = 1}^{20} (n^3 - n)\).

Answers

  1. Find the value that the terms tend towards as n approaches ∞.
  2. –4
  3. does not exist
  4. \(-\frac{1}{2}\)
  5. \(\frac{1}{6}\)
  6. does not exist; diverges
  7. \(-\frac{1}{3}\); converges
  8. \(\frac{5}{2}\); converges
  9. \(\frac{20}{3}\); converges
  10. \(\frac{5}{2}\); converges
  11. \(f′(x) = 6x^2 - 3\)
  12. \(f′(x) = \frac{2}{\sqrt{x}}\)
  13. \(-\frac{1}{4}\)
  14. 1
  15. 43890