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Connecting Infinite Limits and Vertical Asymptotes

Lesson ~10 min read

In simple terms: In simple terms, infinite limits describe when a graph shoots straight up or down towards infinity, which we see as a vertical wall called a vertical asymptote.

Why this matters

Imagine you're at a concert in downtown Chicago. The sound engineer, Maya, is tweaking the levels. She turns a knob for the guitar feedback. As she gets closer and closer to a certain setting, a high-pitched whine starts. It gets louder... and louder... and LOUDER, until it feels like it's going to explode into an infinitely loud screech. The sound is becoming unbounded.

In calculus, we have a way to describe this exact moment of "exploding" behavior. We're not interested in the knob's final position, but what happens as we get infinitely close to it. This idea of a function's output rocketing towards infinity as the input approaches a specific number is the key to understanding infinite limits and their graphical partners, vertical asymptotes.

Concept overview

flowchart TD
    A[Start with f(x) and a point x=c] --> B{Is the denominator of f(x) zero at x=c?};
    B -- No --> C[No Vertical Asymptote at x=c];
    B -- Yes --> D{Is the numerator of f(x) also zero at x=c?};
    D -- No --> E[x=c is a Vertical Asymptote];
    D -- Yes --> F[Simplify f(x) by factoring and canceling];
    F --> G{Does the factor (x-c) still remain in the denominator?};
    G -- Yes --> E;
    G -- No --> H[x=c is a Removable Discontinuity (Hole)];

Core explanation

Hello there. I'm Saavi, and I'm here to help you connect some important dots in calculus. Today, we're talking about what happens when functions go wild.

What is an "Infinite Limit"?

First, let's get our terms straight. An infinite limit is a situation where the y-values of a function grow infinitely large (or infinitely negative) as the x-value gets closer and closer to some number, let's call it c.

Let's look at the classic example: f(x) = 1/x.

What happens as x gets super close to 0? Let's try plugging in some numbers from the right side of 0.

x (approaching 0 from the right) f(x) = 1/x
0.1 10
0.01 100
0.001 1000
0.000001 1,000,000

As x gets closer to 0, f(x) gets enormous. It's heading towards positive infinity. We write this using limit notation:

lim_(x→0⁺) 1/x = ∞

That little + sign on the 0 means "from the right."

Now, what about from the left side of 0?

x (approaching 0 from the left) f(x) = 1/x
-0.1 -10
-0.01 -100
-0.001 -1000
-0.000001 -1,000,000

Here, f(x) is heading towards negative infinity. We write:

lim_(x→0⁻) 1/x = -∞

Connecting Limits to Vertical Asymptotes

So what does this look like on a graph? It looks like a vertical asymptote.

Think of a vertical asymptote as an invisible, electrified fence on a graph. The function can get closer and closer to this fence, but it can never touch it. As it approaches, it gets a massive jolt, sending it shooting straight up or straight down.

Here is the official connection, and you should treat this as a core definition:

The line x = c is a vertical asymptote for a function f(x) if at least one of the following is true:

  • lim_(x→c⁺) f(x) = ∞
  • lim_(x→c⁺) f(x) = -∞
  • lim_(x→c⁻) f(x) = ∞
  • lim_(x→c⁻) f(x) = -∞

If the limit from either the left or the right (or both) goes to ±∞, you've found a vertical asymptote.

How to Find Vertical Asymptotes

For rational functions (a polynomial divided by another polynomial), vertical asymptotes are hiding in one specific place: where the denominator is zero.

Let's say you have a function f(x) = N(x) / D(x).

  1. 1
    Find the candidates
    Set the denominator D(x) equal to zero and solve for x. These are your potential vertical asymptotes. Let's say x = c is one of these values.
  2. 2
    Check the numerator
    Now, plug c into the numerator, N(x).
    • If N(c) is not zero, you have a (non-zero number) / 0 situation. Congratulations, you've found a vertical asymptote at x = c.
    • If N(c) is also zero, you have a 0/0 situation. This is an indeterminate form. It means more investigation is needed.

What to do with 0/0

When you get 0/0, it usually means there's a common factor in the numerator and denominator that can be canceled out.

  • If the factor that caused the zero in the denominator cancels out completely, you have a hole in the graph, not a vertical asymptote.
  • If the factor still remains in the denominator after canceling, you have a vertical asymptote.

Think of it like this: A (non-zero)/0 form is a loud, clear signal of an infinite limit. A 0/0 form is a whisper that you need to lean in closer to understand, by factoring and simplifying.

We'll walk through this exact scenario in the examples below. It's a classic AP exam question, so let's make sure you're ready for it.

Worked examples

Let's put this into practice. The best way to learn this is by doing it.

Example 1

A Straightforward Asymptote

Problem: Find the vertical asymptote(s) for the function f(x) = (x + 5) / (x - 2) and describe the behavior at the asymptote using limits.

Solution Walkthrough:

  1. 1
    Find the candidate
    A vertical asymptote can only happen where the denominator is zero. x - 2 = 0 x = 2 So, our only candidate is the line x = 2.
  2. 2
    Check the numerator
    Now, we check if the numerator is also zero at x = 2. x + 5 at x = 2 is 2 + 5 = 7. Since the numerator is 7 (not zero) when the denominator is zero, we have a 7/0 situation. This confirms that x = 2 is a vertical asymptote.
  3. 3
    Describe the behavior with limits
    Now we need to figure out if the graph goes to +∞ or -∞ on each side of the asymptote. We'll test values very close to 2.
    • From the right (x → 2⁺)
      Let's pick a number just bigger than 2, like x = 2.1. f(2.1) = (2.1 + 5) / (2.1 - 2) = 7.1 / 0.1 = 71 The result is a large positive number. This tells us the function is shooting up to positive infinity. Answer: lim_(x→2⁺) f(x) = ∞
    • From the left (x → 2⁻)
      Let's pick a number just smaller than 2, like x = 1.9. f(1.9) = (1.9 + 5) / (1.9 - 2) = 6.9 / -0.1 = -69 The result is a large negative number. This tells us the function is diving down to negative infinity. Answer: lim_(x→2⁻) f(x) = -∞
Example 2

Asymptote vs. Hole

Problem: Find and classify all discontinuities for g(x) = (x² - 9) / (x² + x - 12).

Solution Walkthrough:

  1. 1
    Factor everything first
    This is always the best first step with rational functions.
    • Numerator: x² - 9 = (x - 3)(x + 3)
    • Denominator: x² + x - 12 = (x + 4)(x - 3) So, g(x) = ((x - 3)(x + 3)) / ((x + 4)(x - 3))
  2. 2
    Find the discontinuities
    The denominator is zero when x = -4 or x = 3. These are our two points of discontinuity.
  3. 3
    Analyze each point
    • At x = 3
      Notice the (x - 3) factor appears in both the top and bottom. We can cancel it! g(x) = (x + 3) / (x + 4), for x ≠ 3. Because we could cancel the factor, x = 3 is a hole, not a vertical asymptote.
    • At x = -4
      The (x + 4) factor is left in the denominator after simplifying. This means we have a (non-zero)/0 situation here. Let's check the simplified function: As x → -4, the numerator approaches -4 + 3 = -1. The denominator approaches 0. This is a (-1)/0 situation. This confirms that x = -4 is a vertical asymptote.

Final Answer: The function g(x) has a vertical asymptote at x = -4 and a hole at x = 3.

Try it yourself

Ready to try a couple on your own? Don't worry about getting the perfect answer right away. Focus on setting up the problem correctly.

  1. 1
    Problem
    Consider the function h(x) = (2x) / (x² - 25). Find all vertical asymptotes. For each one, use one-sided limits to describe the behavior of the function.
    • Hint: Start by factoring the denominator. You should find two separate vertical lines to investigate.
  2. 2
    Problem
    Carlos is analyzing the function k(x) = (x - 4) / (x² - 16). He says there's a vertical asymptote at x = 4. Is he correct?
    • Hint: What happens when you plug x = 4 into the numerator and the denominator? Factor everything and see what happens. Is it a wall (asymptote) or a missing step (hole)?
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