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Infinite Limits and Vertical Asymptotes

When functions shoot off to infinity at a specific point

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Infinite Limits

Sometimes a function doesn't approach a finite value - it shoots off to infinity!

The Notation

lim⁡x→af(x)=∞\lim_{x \to a} f(x) = \infty

This means: as x approaches a, f(x) grows without bound (gets arbitrarily large).

lim⁡x→af(x)=−∞\lim_{x \to a} f(x) = -\infty

This means: as x approaches a, f(x) becomes arbitrarily negative.

Important: When we write =∞= \infty, the limit does not exist (DNE) in the traditional sense. We're just being specific about how it doesn't exist.

Vertical Asymptotes

If lim⁡x→af(x)=±∞\lim_{x \to a} f(x) = \pm\infty, then x = a is a vertical asymptote.

The graph shoots up or down near this vertical line.

Common Cause: Division by Zero

The most common way to get infinite limits: denominator approaches 0 while numerator doesn't.

Example: f(x)=1x−2f(x) = \frac{1}{x - 2}

As x→2x \to 2:

  • Numerator: 1 (constant)
  • Denominator: x−2→0x - 2 \to 0

Result: 1tiny number=huge number\frac{1}{\text{tiny number}} = \text{huge number}

One-Sided Infinite Limits

We often need to check both sides because they might go different directions!

Example: f(x)=1x−2f(x) = \frac{1}{x - 2}

From the left (x→2−x \to 2^-):

  • When x = 1.9: 11.9−2=1−0.1=−10\frac{1}{1.9 - 2} = \frac{1}{-0.1} = -10
  • When x = 1.99: 11.99−2=1−0.01=−100\frac{1}{1.99 - 2} = \frac{1}{-0.01} = -100
  • When x = 1.999: 1−0.001=−1000\frac{1}{-0.001} = -1000

lim⁡x→2−1x−2=−∞\lim_{x \to 2^-} \frac{1}{x - 2} = -\infty

From the right (x→2+x \to 2^+):

  • When x = 2.1: 12.1−2=10.1=10\frac{1}{2.1 - 2} = \frac{1}{0.1} = 10
  • When x = 2.01: 10.01=100\frac{1}{0.01} = 100
  • When x = 2.001: 10.001=1000\frac{1}{0.001} = 1000

lim⁡x→2+1x−2=+∞\lim_{x \to 2^+} \frac{1}{x - 2} = +\infty

The Sign Test

To determine if the limit is +∞+\infty or −∞-\infty:

  1. Find where denominator = 0
  2. Check a test point just to the left
  3. Check a test point just to the right
  4. Determine the sign of the function

Visual Interpretation

On a graph:

  • Vertical asymptote: Dashed vertical line at x = a
  • Left side: Graph shoots up (+∞+\infty) or down (−∞-\infty)
  • Right side: Graph shoots up or down
  • The graph never touches the vertical asymptote

Example with Factoring

f(x)=x+1(x−3)2f(x) = \frac{x + 1}{(x - 3)^2}

As x→3x \to 3:

  • Numerator: 3+1=43 + 1 = 4 (positive)
  • Denominator: (3−3)2=0(3 - 3)^2 = 0

But (x−3)2(x - 3)^2 is always positive (it's squared!)

So from both sides: positivepositive tiny=+∞\frac{\text{positive}}{\text{positive tiny}} = +\infty

lim⁡x→3x+1(x−3)2=+∞\lim_{x \to 3} \frac{x + 1}{(x - 3)^2} = +\infty

Both sides go to +∞+\infty!

📚 Practice Problems

1Problem 1medium

❓ Question:

Find lim⁡x→1−2x−1\lim_{x \to 1^-} \frac{2}{x - 1} and lim⁡x→1+2x−1\lim_{x \to 1^+} \frac{2}{x - 1}

💡 Show Solution

Approaching from the left (x→1−x \to 1^-):

Test with x = 0.9: 20.9−1=2−0.1=−20\frac{2}{0.9 - 1} = \frac{2}{-0.1} = -20

The numerator is positive (2), the denominator is negative (approaching 0 from below).

lim⁡x→1−2x−1=−∞\lim_{x \to 1^-} \frac{2}{x - 1} = -\infty

Approaching from the right (x→1+x \to 1^+):

Test with x = 1.1: 21.1−1=20.1=20\frac{2}{1.1 - 1} = \frac{2}{0.1} = 20

The numerator is positive (2), the denominator is positive (approaching 0 from above).

lim⁡x→1+2x−1=+∞\lim_{x \to 1^+} \frac{2}{x - 1} = +\infty

Conclusion:

  • Left-hand limit: −∞-\infty
  • Right-hand limit: +∞+\infty
  • There is a vertical asymptote at x = 1

2Problem 2hard

❓ Question:

Find lim⁡x→−2x(x+2)2\lim_{x \to -2} \frac{x}{(x + 2)^2} and describe the behavior

💡 Show Solution

Step 1: Check what happens at x = -2

Numerator: x→−2x \to -2 (negative) Denominator: (x+2)2→0(x + 2)^2 \to 0 (positive, because it's squared)

Step 2: Determine the sign

negativepositive tiny=large negative\frac{\text{negative}}{\text{positive tiny}} = \text{large negative}

Step 3: Check both sides

Since (x+2)2(x + 2)^2 is always positive (squared term), and the numerator x is negative near -2, the function will be negative on both sides.

From the left (x = -2.1): −2.1(−2.1+2)2=−2.10.01=−210\frac{-2.1}{(-2.1 + 2)^2} = \frac{-2.1}{0.01} = -210

From the right (x = -1.9): −1.9(−1.9+2)2=−1.90.01=−190\frac{-1.9}{(-1.9 + 2)^2} = \frac{-1.9}{0.01} = -190

Both sides go to −∞-\infty!

lim⁡x→−2x(x+2)2=−∞\lim_{x \to -2} \frac{x}{(x + 2)^2} = -\infty

Behavior: There is a vertical asymptote at x = -2, and the graph approaches −∞-\infty from both sides.

3Problem 3easy

❓ Question:

Find lim(x→3⁺) 1/(x - 3)²

💡 Show Solution

Step 1: Analyze what happens as x→3⁺: x approaches 3 from the right (x > 3)

Step 2: Examine denominator: x - 3 approaches 0 from the positive side (x - 3)² is always positive and approaches 0

Step 3: Analyze the fraction: 1/(small positive number) = large positive number As denominator→0⁺, fraction→+∞

Step 4: Test values: x = 3.1: 1/(0.1)² = 1/0.01 = 100 x = 3.01: 1/(0.01)² = 10,000 Pattern: approaching +∞

Step 5: Conclusion: lim(x→3⁺) 1/(x - 3)² = +∞

Answer: +∞

4Problem 4medium

❓ Question:

Determine lim(x→-2⁻) (x + 1)/(x + 2)

💡 Show Solution

Step 1: Analyze as x→-2⁻: Approaching -2 from the left (x < -2)

Step 2: Evaluate numerator: x + 1 → -2 + 1 = -1 (negative)

Step 3: Evaluate denominator: x + 2 approaches 0 from the left (negative side)

Step 4: Determine sign: (negative)/(small negative) = large positive lim(x→-2⁻) (x + 1)/(x + 2) = +∞

Step 5: Verify with values: x = -2.1: (-1.1)/(-0.1) = 11 (positive) x = -2.01: (-1.01)/(-0.01) = 101 Approaching +∞ ✓

Answer: +∞

5Problem 5hard

❓ Question:

Find the vertical asymptote(s) of f(x) = (x - 1)/(x² - 4) and determine the behavior near each.

💡 Show Solution

Step 1: Find vertical asymptotes: Set denominator = 0 x² - 4 = 0 x = ±2

Step 2: Check if numerator ≠ 0 at these points: At x = 2: numerator = 2 - 1 = 1 ≠ 0 ✓ At x = -2: numerator = -2 - 1 = -3 ≠ 0 ✓ Both are vertical asymptotes

Step 3: Analyze behavior near x = 2: lim(x→2⁻): (positive)/(small negative) = -∞ lim(x→2⁺): (positive)/(small positive) = +∞

Step 4: Analyze behavior near x = -2: lim(x→-2⁻): (negative)/(small positive) = -∞ lim(x→-2⁺): (negative)/(small negative) = +∞

Answer: Vertical asymptotes at x = ±2 x = 2: -∞ from left, +∞ from right x = -2: -∞ from left, +∞ from right

Explain using:

📋 AP Calculus AB — Exam Format Guide

⏱ 3 hours 15 minutes📝 51 questions📊 4 sections
SectionFormatQuestionsTimeWeightCalculator
Multiple Choice (No Calculator)MCQ3060 min33.3%🚫
Multiple Choice (Calculator)MCQ1545 min16.7%✅
Free Response (Calculator)FRQ230 min16.7%✅
Free Response (No Calculator)FRQ460 min33.3%🚫

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💡 Key Test-Day Tips

  • ✓Show all work on FRQs
  • ✓Use proper notation
  • ✓Check units
  • ✓Manage your time

⚠️ Common Mistakes: Infinite Limits and Vertical Asymptotes

Avoid these 4 frequent errors

🌍 Real-World Applications: Infinite Limits and Vertical Asymptotes

See how this math is used in the real world

📝 Worked Example: Related Rates — Expanding Circle

Problem:

A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 22 cm/s. How fast is the area of the circle increasing when the radius is 1010 cm?

2Write the relationship between variables
3Differentiate both sides with respect to time
4Substitute known values
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❓ Frequently Asked Questions

What is Infinite Limits and Vertical Asymptotes?▾
When functions shoot off to infinity at a specific point
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.