NOTE: You may view the original PDFs of the notes written in class by clicking on the link for the 1:30 Section or 2:30 Section, respectively.
Lecture 21: Limits at Infinity
Goal: Understand the end behavior (asymptotic behavior) of functions.
Definition. We say:
\[ \displaystyle\lim_{x \to \infty} f(x) = L \]to mean \( f(x) \) goes to \( L \) as \( x \) "goes to infinity" (i.e., \( x \) gets very, very large). Similarly,
\[ \displaystyle\lim_{x \to -\infty} f(x) = L \]means \( f(x) \) goes to \( L \) as \( x \) "goes to negative infinity" (i.e., \( x \) gets very, very "largely negative").
NOTE: A number being "small" typically refers to when it is close to 0. Hence, why I am using the phrase "largely negative," even though it is a bit awkward.
Example 1.
For each part, determine \( \displaystyle\lim_{x \to \infty} f(x) \) and \( \displaystyle\lim_{x \to -\infty} f(x) \), where the graph of \( f \) is given:
Figure 1
\( \displaystyle\lim_{x \to \infty} f(x) = \) 1
\( \displaystyle\lim_{x \to -\infty} f(x) = \) 0
Figure 2
\( \displaystyle\lim_{x \to \infty} f(x) = \) \( \infty \)
\( \displaystyle\lim_{x \to -\infty} f(x) = \) \( -\infty \)
Example 2.
Determine the end behavior (limits at infinity) for the function \( f(x) = \frac{1}{x} \)
Figure 3
| \( x \) | \( f(x) \) |
|---|---|
| 1 | 1 |
| 100 | 0.01 |
| 10000 | 0.0001 |
| 1000000 | 0.000001 |
| \( x \) | \( f(x) \) |
|---|---|
| -1 | -1 |
| -100 | -0.01 |
| -10000 | -0.0001 |
| -1000000 | -0.000001 |
In general, for any positive integer \( n \),
\[ \displaystyle\lim_{x \to \infty} \frac{1}{x^n} = \displaystyle\lim_{x \to -\infty} \frac{1}{x^n} = 0 \]Example 3.
Compute \( \displaystyle\lim_{x \to \infty} \left( 2 + \frac{3}{x^2} \right) \)
\[ \begin{gathered} \displaystyle\lim_{x \to \infty} \left( 2 + \frac{3}{x^2} \right) = \displaystyle\lim_{x \to \infty} 2 + 3 \lim_{x \to \infty} \frac{1}{x^2} \\[0.5em] = 2 + 3 \cdot 0 \\[0.5em] = 2 \end{gathered} \]Rational Functions
Main Strategy (Formally): Factor out the highest power of \( x \) in the denominator.
Example 4.
For \( f(x) = \frac{x^2 - 1}{2x^2 + 1} \), compute \( \displaystyle\lim_{x \to \infty} f(x) \) and \( \displaystyle\lim_{x \to -\infty} f(x) \).
\[ \begin{gathered} \displaystyle\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^2\left(\frac{x^2}{x^2} - \frac{1}{x^2}\right)}{x^2\left(\frac{2x^2}{x^2} + \frac{1}{x^2}\right)} = \displaystyle\lim_{x \to \infty} \frac{1 - \frac{1}{x^2}}{2 + \frac{1}{x^2}} = \frac{1 - 0}{2 + 0} = \frac{1}{2} \\[0.5em] \displaystyle\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{1 - \frac{1}{x^2}}{2 + \frac{1}{x^2}} = \frac{1}{2} \end{gathered} \]Example 5.
For \( f(x) = \frac{x^2 + 2}{x^3 + x^2 - 1} \), compute \( \displaystyle\lim_{x \to \infty} f(x) \) and \( \displaystyle\lim_{x \to -\infty} f(x) \).
\[ \begin{gathered} \displaystyle\lim_{x \to \infty} \frac{x^2 + 2}{x^3 + x^2 - 1} = \displaystyle\lim_{x \to \infty} \frac{x^3\left(\frac{x^2}{x^3} + \frac{2}{x^3}\right)}{x^3\left(\frac{x^3}{x^3} + \frac{x^2}{x^3} - \frac{1}{x^3}\right)} \\[0.5em] = \lim_{x \to \infty} \frac{\frac{1}{x} + \frac{2}{x^3}}{1 + \frac{1}{x} - \frac{1}{x^3}} = \frac{0 + 0}{1 + 0 - 0} = 0 \\[0.5em] \displaystyle\lim_{x \to -\infty} \frac{x^2 + 2}{x^3 + x^2 - 1} = \displaystyle\lim_{x \to -\infty} \frac{\frac{1}{x} + \frac{2}{x^3}}{1 + \frac{1}{x} - \frac{1}{x^3}} = 0 \end{gathered} \]Main Strategy (Informally): Compare the leading terms
Example 6.
For \( f(x) = \frac{x^2 + x}{-x + 3} \), compute \( \displaystyle\lim_{x \to \infty} f(x) \) and \( \displaystyle\lim_{x \to -\infty} f(x) \).
\[ \begin{gathered} \displaystyle\lim_{x \to \infty} \frac{x^2 + x}{-x + 3} = \displaystyle\lim_{x \to \infty} \frac{x^2}{-x} = \lim_{x \to \infty} (-x) = -\infty \\[0.5em] \displaystyle\lim_{x \to -\infty} \frac{x^2 + x}{-x + 3} = \displaystyle\lim_{x \to -\infty} \frac{x^2}{-x} = \displaystyle\lim_{x \to -\infty} (-x) = \infty \end{gathered} \]In Summary: For a rational function:
\[ f(x) = \frac{a(x)}{b(x)} \]if \( LT(a(x)) \) and \( LC(a(x)) \) denote the leading term and leading coefficient of the polynomial \( a(x) \) respectively, then,
\[ \begin{gathered} \displaystyle\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{LT(a(x))}{LT(b(x))} \\[0.5em] \lim_{x \to -\infty} f(x) = \displaystyle\lim_{x \to -\infty} \frac{LT(a(x))}{LT(b(x))} \end{gathered} \]In particular,
- If \( \deg(a(x)) = \deg(b(x)) \), then: \[ \lim_{x \to \infty} f(x) = \displaystyle\lim_{x \to -\infty} f(x) = \frac{LC(a(x))}{LC(b(x))} \]
- If \( \deg(a(x)) < \deg(b(x)) \), then: \[ \lim_{x \to \infty} f(x) = \displaystyle\lim_{x \to -\infty} f(x) = 0 \]
- If \( \deg(a(x)) > \deg(b(x)) \), more work is needed (since it depends on whether \( \deg(a(x)) - \deg(b(x)) \) is even or odd as well as whether \( \frac{LC(a(x))}{LC(b(x))} \) is positive or negative).
Horizontal and Vertical Asymptotes (HAs and VAs)
Definition. A function \( f(x) \) has a vertical asymptote at \( x = c \) if either \( \displaystyle\lim_{x \to c^-} f(x) \) or \( \displaystyle\lim_{x \to c^+} f(x) \) is NOT finite.
Definition. A function \( f(x) \) has a horizontal asymptote at \( y = L \) if either \( \displaystyle\lim_{x \to -\infty} f(x) = L \) or \( \displaystyle\lim_{x \to \infty} f(x) = L \).
Example 7.
Find the locations of any Horizontal and Vertical Asymptotes for \( f(x) = \frac{3x^2 + 5}{x^2 - 4} \)
For a rational function, VAs occur when the denominator is 0, but the numerator is non-zero (after canceling out any common factors on the top and bottom).
\[ \begin{gathered} x^2 - 4 \stackrel{\text{set}}{=} 0 \\[0.5em] \Rightarrow \quad x = 2, -2 \end{gathered} \]The numerator is not \( 0 \) at \( x = 2 \) and \( x = -2 \), so \( f \) has VAs at \( x = -2 \) and \( x = 2 \).
For a rational function \( f(x) \), HAs occur when either \( \displaystyle\lim_{x \to -\infty} f(x) \) or \( \lim_{x \to \infty} f(x) \) is finite.
NOTE:
Rational functions have at most one HA.
\( \displaystyle\lim_{x \to -\infty} f(x) \) is also \( 3 \)
Slant (Oblique) Asymptotes
Recall Polynomial Long Division:
Theorem (Polynomial Division Algorithm). For a rational function \( \frac{a(x)}{b(x)} \) there are unique polynomials \( q(x) \) (called the quotient) and \( r(x) \) (called the remainder) such that:
\[ \frac{a(x)}{b(x)} = q(x) + \frac{r(x)}{b(x)} \quad \text{AND} \quad \deg(r(x)) < \deg(b(x)) \]Compare this with numbers:
\[ \begin{array}{r} 3 \phantom{)} \\ 2\overline{)\,7\phantom{0}} \\ \underline{-\,6\phantom{0}} \\ 1\phantom{0} \end{array} \]So \( 7 = 3(2) + 1 \), or
\[ \frac{7}{2} = 3 + \frac{1}{2} \]Example 8.
Find the quotient and remainder when \( 2x^2 + 3x + 1 \) is divided by \( x - 1 \).
\[ \begin{array}{r} 2x + 5 \phantom{)} \\ x-1\overline{)\,2x^2 + 3x + 1} \\ \underline{-\,(2x^2 - 2x)\phantom{+1}} \\ 5x + 1 \\ \underline{-\,(5x - 5)} \\ 6 \end{array} \]We stop long division since
\[ \deg(6) = 0 < 1 = \deg(x-1) \]Thus,
\[ \frac{2x^2 + 3x + 1}{x - 1} = (2x + 5) + \frac{6}{x - 1} \]Definition. The line \( y = mx + b \) is a slant (oblique) asymptote of the graph of \( f(x) \) if:
\[ \displaystyle\lim_{x \to \infty} [f(x) - (mx + b)] = 0 \quad \text{OR} \quad \displaystyle\lim_{x \to -\infty} [f(x) - (mx + b)] = 0 \]Informally, this means the graph of \( f(x) \) gets closer and closer to the line \( y = mx + b \) as \( x \) goes to positive or negative infinity.
Theorem. The following are equivalent for a rational function \( \frac{a(x)}{b(x)} \):
- \( \frac{a(x)}{b(x)} \) has a slant asymptote.
- The quotient in polynomial long division is a degree 1 polynomial.
- \( \deg(a(x)) = \deg(b(x)) + 1 \)
If these hold, the quotient obtained from long division is a slant asymptote.
Example 9.
Does the function \( f(x) = \frac{2x^2 + 3x + 1}{x - 1} \) have a slant asymptote? If so, find it.
Yes! Since the degree of the denominator (1) is exactly one less than the degree of the numerator (2). Moreover, Perform the long division we get
\[ \frac{2x^2 + 3x + 1}{x - 1} = [2x + 5] - \frac{6}{x - 1} \]So \( f \) has \( y = 2x + 5 \) as a slant asymptote. We can see this in the graph of \( f(x) \); since \( -\frac{6}{x-1} \rightarrow 0 \) as \( x \to \pm\infty \), \( f(x) \) starts getting closer to the value of \( 2x+5 \) as \( x \to \pm\infty \)
Figure 4
Example 10.
Find all asymptotes (HAs, VAs, and slant) for the function
\[ f(x) = \frac{2x^3 - 3x + 1}{x^2 - x - 6} \]Vertical Asymptotes:
\[ \begin{gathered} x^2 - x - 6 \stackrel{\text{set}}{=} 0 \\[0.5em] (x-3)(x+2) = 0 \\[0.5em] \Longrightarrow x = -2, 3 \end{gathered} \]\( \) \( 2x^3 - 3x + 1 \) is not zero at \( x = -2 \) nor at \( x = 3 \), so those are the locations of the VAs.
Horizontal Asymptotes:
Comparing the degrees, \( f \) has a slant asymptote. So, (since \( f \) is rational) \( f \) can not have a HA.
Slant Asymptotes:
Performing the long division.
\[ \begin{array}{r} 2x+2 \phantom{\quad} \\[-2pt] x^2-x-6 \overline{\smash{)}\, 2x^3-3x+1\phantom{)}} \\[-2pt] \underline{-\,(2x^3-2x^2-12x)\phantom{)}} \\[-2pt] 2x^2+9x+1\phantom{)} \\[-2pt] \underline{-\quad (2x^2-2x-12)\phantom{)}} \\[-2pt] 11x+13\phantom{)} \end{array} \]We get that \( f \) has \( y = 2x+2 \) as a slant asymptote.
