Infinite Limits and Vertical Asymptotes

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 $\lim_{x \to a} f(x) = \pm\infty$, 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: $f(x) = \frac{1}{x - 2}$

As $x \to 2$:

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

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

One-Sided Infinite Limits

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

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

From the left ($x \to 2^-$):

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

$\lim_{x \to 2^-} \frac{1}{x - 2} = -\infty$

From the right ($x \to 2^+$):

• When x = 2.1: $\frac{1}{2.1 - 2} = \frac{1}{0.1} = 10$
• When x = 2.01: $\frac{1}{0.01} = 100$
• When x = 2.001: $\frac{1}{0.001} = 1000$

$\lim_{x \to 2^+} \frac{1}{x - 2} = +\infty$

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

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

As $x \to 3$:

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

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

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

$\lim_{x \to 3} \frac{x + 1}{(x - 3)^2} = +\infty$

Both sides go to $+\infty$!

Step 2: Determine the sign

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

Step 3: Check both sides

Since $(x + 2)^2$ 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): $\frac{-2.1}{(-2.1 + 2)^2} = \frac{-2.1}{0.01} = -210$

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

Both sides go to $-\infty$!

$\lim_{x \to -2} \frac{x}{(x + 2)^2} = -\infty$

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