Infinite Limits for MATH 139

What is an Infinite Limit?

An infinite limit happens when a function grows without bound (gets infinitely large or infinitely negative) as $x$ approaches some value.

Notation:

• $\displaystyle\lim_{x \to a} f(x) = \infty$ means $f(x)$ grows toward positive infinity
• $\displaystyle\lim_{x \to a} f(x) = -\infty$ means $f(x)$ grows toward negative infinity

Important: Technically, the limit "does not exist" when it's $\pm\infty$, because infinity isn't a real number. But we write $= \infty$ to describe the behavior.

In this graph, we see the function $f(x) = \frac{1}{x}$. Notice how the function shoots up toward $+\infty$ as $x$ approaches $0$ from the right, and plunges down toward $-\infty$ as $x$ approaches $0$ from the left. The y-axis ($x = 0$) is a vertical asymptote.

Vertical Asymptotes

When $\displaystyle\lim_{x \to a} f(x) = \pm\infty$, the line $x = a$ is called a vertical asymptote.

The graph shoots up or down toward this vertical line but never actually touches it.

When Do Infinite Limits Happen?

Infinite limits occur when:

1. The denominator approaches 0
2. The numerator approaches a non-zero number

$$\frac{\text{non-zero}}{\text{something approaching } 0} \to \pm\infty$$

How to Determine the Sign ($+\infty$ or $-\infty$)

The key question: What's the sign of the function near $x = a$?

Check:

1. Is the numerator positive or negative near $a$?
2. Is the denominator approaching $0^+$ (small positive) or $0^-$ (small negative)?

$$\frac{(+)}{(0^+)} = +\infty \qquad \frac{(+)}{(0^-)} = -\infty$$ $$\frac{(-)}{(0^+)} = -\infty \qquad \frac{(-)}{(0^-)} = +\infty$$

In this graph, we see $f(x) = \frac{1}{x-2}$ with a vertical asymptote at $x = 2$. From the left ($x \to 2^-$), the function goes to $-\infty$. From the right ($x \to 2^+$), it goes to $+\infty$. This is why the two-sided limit DNE.

Example 4: Hole vs Asymptote

Problem: Analyze $f(x) = \frac{x^2-4}{x-2}$

Show Solution

Step 1: Factor: $$\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x + 2 \quad (x \neq 2)$$

Step 2: At $x = 2$, both numerator and denominator equal zero.

After canceling, we get $x + 2$, which is defined at $x = 2$.

This means $x = 2$ is a hole (removable discontinuity), NOT a vertical asymptote!

$$\lim_{x \to 2} \frac{x^2-4}{x-2} = 2 + 2 = \boxed{4}$$

In this graph, compare two functions: $f(x) = \frac{1}{x-2}$ (which has a vertical asymptote at $x=2$) and $g(x) = \frac{x^2-4}{x-2}$ (which simplifies to the line $y = x+2$ with a hole at $(2, 4)$). Notice the dramatic difference — one blows up to infinity, the other just has a single missing point!

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Key Distinction

Common Mistakes and Misunderstandings

❌ Mistake: Writing $\lim = \infty$ means the limit "exists"

Wrong thinking: "The limit equals infinity, so it exists."

Why it's wrong: Infinity is not a real number. When we write $\lim_{x \to a} f(x) = \infty$, we're describing behavior, not stating an actual value. The limit technically does not exist (DNE).

Correct: Say "the limit is infinite" or "the limit does not exist (it goes to infinity)."

❌ Mistake: Confusing $\frac{0}{0}$ with $\frac{\text{non-zero}}{0}$

Wrong: "The denominator is zero, so the limit is $\infty$."

Why it's wrong:

• $\frac{0}{0}$ is indeterminate — you need to simplify first (often gives a finite limit or a hole)
• $\frac{\text{non-zero}}{0}$ gives $\pm\infty$ (vertical asymptote)

Correct: Always check BOTH the numerator AND denominator before concluding.

❌ Mistake: Forgetting to check one-sided limits

Wrong: $\displaystyle\lim_{x \to 2} \frac{1}{x-2} = \infty$

Why it's wrong: From the left, it goes to $-\infty$. From the right, it goes to $+\infty$. Since they disagree, the two-sided limit DNE (not just "equals infinity").

Correct: $$\lim_{x \to 2^-} \frac{1}{x-2} = -\infty, \quad \lim_{x \to 2^+} \frac{1}{x-2} = +\infty$$ $$\therefore \lim_{x \to 2} \frac{1}{x-2} \text{ DNE}$$

❌ Mistake: Assuming every zero in the denominator is a vertical asymptote

Wrong: "$f(x) = \frac{x-3}{x-3}$ has a vertical asymptote at $x = 3$."

Why it's wrong: After canceling, $f(x) = 1$ for all $x \neq 3$. There's a hole at $x = 3$, not an asymptote.

Correct: Only $\frac{\text{non-zero}}{0}$ creates a vertical asymptote. If both are zero, factor and simplify first.

❌ Mistake: Getting the sign wrong on one-sided limits

Wrong: $\displaystyle\lim_{x \to 5^-} \frac{3}{x-5} = +\infty$

Why it's wrong: When $x < 5$, we have $x - 5 < 0$, so the denominator is negative. $$\frac{(+)}{(0^-)} = -\infty$$

Correct: $\displaystyle\lim_{x \to 5^-} \frac{3}{x-5} = -\infty$

Tip: Plug in a test value! Try $x = 4.9$: $\frac{3}{4.9-5} = \frac{3}{-0.1} = -30$ (negative!)