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Infinite Limits &
Vertical Asymptotes

Explore how functions behave near points where they blow up to \(\pm\infty\), and how this connects to vertical asymptotes on their graphs.

AP Calculus AB

What Is an Infinite Limit?

An infinite limit describes the behavior of a function \(f(x)\) whose output grows without bound as \(x\) approaches some finite value \(c\). We write:

Technically, the limit does not exist as a real number: we use the \(\infty\) notation to describe how it fails to exist.

Visualizing Infinite Limits

\(x = 2\) x y 2 0 \(-\infty\) \(+\infty\)

What Is a Vertical Asymptote?

A vertical asymptote is a vertical line \(x = c\) that the graph of a function approaches but never crosses (in a neighborhood of \(c\)). The curve shoots off toward \(\pm\infty\) near this line.

Graphical View
The graph gets arbitrarily close to the dashed line \(x = c\) without touching it.
Analytic View
At least one of the one-sided limits at \(x = c\) equals \(+\infty\) or \(-\infty\).

The Core Connection

In other words: infinite limits and vertical asymptotes are two descriptions of the same phenomenon: one is the analytic (limit) description, the other is the geometric (graph) description.

Infinite Limit
"The function blows up": \(\lim_{x\to c} f(x) = \pm\infty\)
Vertical Asymptote
"The graph has a vertical wall at \(x = c\)"

One-Sided Behavior Matters

\(x = c\) x y from left: \(+\infty\) from right: \(+\infty\) \(f(x) = \dfrac{1}{(x-c)^2}\)

The two one-sided limits can behave differently. For \(\frac{1}{x-c}\), they go in opposite directions (\(+\infty\) and \(-\infty\)). For \(\frac{1}{(x-c)^2}\), both sides go to \(+\infty\). Either way, there is a vertical asymptote at \(x = c\).

Infinite Limit vs. Limit at Infinity

Infinite Limit Limit at Infinity
Notation \(\lim_{x \to c} f(x) = \pm\infty\) \(\lim_{x \to \infty} f(x) = L\)
What varies \(x\) approaches a finite value \(x\) grows without bound
Output Output → \(\pm\infty\) Output → finite \(L\)
Graph feature Vertical asymptote Horizontal asymptote
Limit exists? No (not a real number) Yes, equals \(L\)

How to Find Vertical Asymptotes

1
Factor the numerator and denominator of \(f(x)\) completely.
2
Cancel any common factors. A cancelled factor creates a hole, not a vertical asymptote.
3
Set the remaining denominator equal to zero and solve for \(x\). Each solution is a vertical asymptote.
4
Confirm by checking that at least one of \(\displaystyle\lim_{x \to c^+} f(x)\) or \(\displaystyle\lim_{x \to c^-} f(x)\) is \(\pm\infty\).

Determining the Sign of Infinity

Once you know \(x = c\) is a vertical asymptote, determine whether \(f(x) \to +\infty\) or \(-\infty\) from each side using a sign analysis:

1
Pick a test value slightly less than \(c\) (e.g., \(c - 0.01\)) and evaluate the sign of each factor in numerator and denominator.
2
Pick a test value slightly greater than \(c\) (e.g., \(c + 0.01\)) and do the same.
3
The overall sign of the fraction tells you: positive → \(+\infty\), negative → \(-\infty\).

Example 1: \(f(x) = \dfrac{1}{x - 3}\)

3 0 \(-\infty\) \(+\infty\) \(x=3\)

Example 2: \(g(x) = \dfrac{x+1}{x^2-1}\)

1
Factor: \(g(x) = \dfrac{x+1}{(x+1)(x-1)}\)
2
Cancel: \(\dfrac{\cancel{x+1}}{\cancel{(x+1)}(x-1)} = \dfrac{1}{x-1}\), for \(x \neq -1\).
3
Result: \(x = -1\) gives a hole (removable discontinuity), not a VA.
4
Remaining denominator: \(x - 1 = 0 \Rightarrow x = 1\) is a vertical asymptote.

Example 3: \(h(x) = \dfrac{2x}{(x+4)^2}\)

\(x=-4\) -4 0

Multiplicity and the Sign of Infinity

The multiplicity (exponent) of the factor in the denominator determines how the function behaves on either side of the asymptote.

Odd Multiplicity
Factor like \((x - c)^1\) or \((x-c)^3\): the sign changes across the asymptote. One side → \(+\infty\), the other → \(-\infty\).
Even Multiplicity
Factor like \((x - c)^2\) or \((x-c)^4\): the sign stays the same on both sides. Both sides → \(+\infty\) or both → \(-\infty\).

Vertical Asymptotes in Special Functions

\(\tan x\)
VAs at \(x = \frac{\pi}{2} + n\pi\) where \(\cos x = 0\).
\(\ln x\)
VA at \(x = 0\): \(\lim_{x\to 0^+} \ln x = -\infty\).
\(\csc x\)
VAs at \(x = n\pi\) where \(\sin x = 0\).

Common Mistakes to Avoid

✕ Denominator = 0 Always Means VA
If the numerator is also zero at \(x = c\), you may have a hole (removable discontinuity) instead. Always factor and cancel first.
✕ Confusing Infinite Limit with Limit at Infinity
\(\lim_{x\to c} f(x) = \infty\) is about \(x\) near a finite value. \(\lim_{x\to\infty} f(x) = L\) is about \(x\) growing unbounded. Different concepts!
✕ Ignoring One-Sided Limits
Stating \(\lim_{x\to c} f(x) = \infty\) without checking both sides. The left and right limits may differ in sign: always verify both.

Summary at a Glance

Scenario Limit Behavior Graph Feature
\(\lim_{x\to c} f(x) = +\infty\) Both sides → \(+\infty\) VA, curve rises on both sides
\(\lim_{x\to c} f(x) = -\infty\) Both sides → \(-\infty\) VA, curve falls on both sides
Left → \(-\infty\), Right → \(+\infty\) Opposite signs VA, curve in opposite directions
Numerator & denom. both zero May be finite after cancellation Hole (removable), not a VA

Question 1 of 5

What is \(\displaystyle\lim_{x \to 5^+} \frac{2}{x - 5}\)?

Question 2 of 5

The function \(f(x) = \dfrac{x - 2}{x^2 - 4}\) has a vertical asymptote at which value?

Question 3 of 5

If the denominator of \(f(x)\) has the factor \((x - 1)^2\) and the numerator is positive near \(x = 1\), what are the one-sided limits?

Question 4 of 5

Which statement correctly describes the relationship between infinite limits and vertical asymptotes?

Question 5 of 5

How many vertical asymptotes does \(f(x) = \dfrac{3x}{x^2 - 9}\) have?