Calculus BC ยท Unit 1
Decide whether a graph runs off a vertical asymptote or settles toward a horizontal asymptote. Watch the sign when x goes to negative infinity.
Jump to section:
A small denominator can send the value to infinity
If the numerator approaches a nonzero number and the denominator approaches 0, look at the sign of the denominator. A positive tiny denominator gives \(\infty\) when the numerator is positive. A negative tiny denominator gives \(-\infty\).
Degrees decide the end behavior of a rational function
For a rational function, compare the degree of the numerator with the degree of the denominator.
Higher degree in the denominator: the limit at infinity is 0.
Equal degrees: the limit is the ratio of the leading coefficients.
Higher degree in the numerator: the limit is infinite. There is no horizontal asymptote.
A square root uses an absolute value
For large \(|x|\), \(\sqrt{x^2 + c} = |x|\sqrt{1 + \tfrac{c}{x^2}}\).
\(|x| = x\) when \(x \to \infty\), and \(|x| = -x\) when \(x \to -\infty\).
Horizontal versus vertical
A finite limit at infinity is a horizontal asymptote. An infinite limit at a finite x-value is a vertical asymptote.
One infinite limit, or the leading-term ratio.
As x approaches 3 from the right, \(x - 3\) is a small positive number.
The numerator is positive.
A positive number divided by a tiny positive number grows without a ceiling.
\(\lim_{x \to 3^+} \dfrac{1}{x - 3} = \infty\)
Answer: \(\infty\)
From the left, \(x - 3\) is a small negative number.
The numerator is positive.
\(\lim_{x \to 3^-} \dfrac{1}{x - 3} = -\infty\)
Answer: \(-\infty\)
\(\lim_{x \to 3^-} \dfrac{1}{x - 3} = -\infty\)
\(\lim_{x \to 3^+} \dfrac{1}{x - 3} = \infty\)
The one-sided limits are not equal.
The two-sided limit does not exist.
Answer: does not exist
The denominator has the higher degree.
Divide the numerator and the denominator by \(x^2\).
\(\dfrac{\tfrac{3}{x} + \tfrac{1}{x^2}}{1 + \tfrac{4}{x^2}}\)
As \(x \to \infty\), the fractions with x in the denominator approach 0.
\(0\)
Answer: \(0\)
The degrees are equal.
The limit is the ratio of the leading coefficients.
\(\dfrac{4}{2} = 2\)
Answer: \(2\)
The degrees are equal, so the same ratio works at \(-\infty\).
\(\dfrac{2}{5}\)
Answer: \(\dfrac{2}{5}\)
Name the asymptote, or compare the degrees.
The degrees are equal.
\(\dfrac{5}{1} = 5\)
The horizontal asymptote is \(y = 5\).
Answer: \(5\)
The numerator has the higher degree.
For large positive x, the fraction is positive and grows without a ceiling.
\(\infty\)
There is no horizontal asymptote.
Answer: \(\infty\)
The degrees are equal.
\(\dfrac{6}{2} = 3\)
Answer: \(3\)
Equal degrees. The leading coefficients control both ends.
\(\dfrac{6}{2} = 3\)
Answer: \(3\)
The numerator approaches 4, which is positive.
From the left, \(x - 2\) is a small negative number.
\(-\infty\)
Answer: \(-\infty\)
Equal degrees: use the leading coefficients.
A has a higher degree in the numerator, so no horizontal asymptote.
B has a higher degree in the denominator, so the horizontal asymptote is \(y = 0\).
C: \(\dfrac{6}{3} = 2\).
D: \(\dfrac{2}{3}\).
Answer: C. \(y = 2\)
A conjugate, a square root at negative infinity, or a rational function with both kinds of asymptotes.
Substitution gives \(\infty - \infty\). Multiply by the conjugate.
\(\dfrac{\bigl(\sqrt{x^2 + 6x} - x\bigr)\bigl(\sqrt{x^2 + 6x} + x\bigr)}{\sqrt{x^2 + 6x} + x}\)
\(\dfrac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} + x} = \dfrac{6x}{\sqrt{x^2 + 6x} + x}\)
For \(x > 0\), divide the numerator and the denominator by \(x\).
\(\sqrt{x^2 + 6x} = x\sqrt{1 + \dfrac{6}{x}}\)
\(\dfrac{6}{\sqrt{1 + \dfrac{6}{x}} + 1}\)
As \(x \to \infty\):
\(\dfrac{6}{1 + 1} = 3\)
Answer: \(3\)
For \(x > 0\),
\(\sqrt{x^2 + 4} = x\sqrt{1 + \dfrac{4}{x^2}}\)
\(\dfrac{2x + 1}{x\sqrt{1 + \dfrac{4}{x^2}}} = \dfrac{2 + \dfrac{1}{x}}{\sqrt{1 + \dfrac{4}{x^2}}}\)
As \(x \to \infty\):
\(\dfrac{2}{1} = 2\)
Answer: \(2\)
For \(x \lt 0\), \(\sqrt{x^2} = |x| = -x\).
\(\sqrt{x^2 + 4} = -x\sqrt{1 + \dfrac{4}{x^2}}\)
\(\dfrac{2x + 1}{-x\sqrt{1 + \dfrac{4}{x^2}}} = \dfrac{2 + \dfrac{1}{x}}{-\sqrt{1 + \dfrac{4}{x^2}}}\)
As \(x \to -\infty\):
\(\dfrac{2}{-1} = -2\)
Answer: \(-2\)
The denominator has degree 2. The numerator has degree 1.
\(\lim_{x \to \infty} f(x) = 0\)
The horizontal asymptote is \(y = 0\).
The numerator is not 0 at \(x = 4\) or at \(x = -2\), so those are vertical asymptotes. They do not change the end behavior.
Answer: \(0\), with horizontal asymptote \(y = 0\)
At \(x = 2\), the denominator is 0 and the numerator is \(12 - 2 = 10\).
That is a vertical asymptote, so the two-sided limit is not 0.
The degrees are equal.
\(\lim_{x \to \infty} f(x) = \dfrac{3}{1} = 3\)
Answer: B. the limit at infinity is 3
For \(x > 0\),
\(\sqrt{4x^2 + 1} = 2x\sqrt{1 + \dfrac{1}{4x^2}}\)
\(\dfrac{8x - 3}{2x\sqrt{1 + \dfrac{1}{4x^2}}} = \dfrac{8 - \dfrac{3}{x}}{2\sqrt{1 + \dfrac{1}{4x^2}}}\)
As \(x \to \infty\):
\(\dfrac{8}{2} = 4\)
Answer: \(4\)
Need help with infinite limits and end behavior?
Book a Session with Amy