Limit Properties for MATH 139

Exam Relevance for MATH 139

Likelihood of appearing: High

Foundation for computing limits. Know sum, product, quotient, and composition rules.

This skill appears on:

Practice Final #1

Lesson

What are Limit Properties?

Limit properties (also called limit laws) are rules that let you break down complicated limits into simpler pieces.

The Idea: Instead of evaluating one big scary limit, you can:

Split it into smaller limits
Evaluate each small limit
Combine the results

These rules work as long as the individual limits exist.

The Basic Limit Properties

Assume $\displaystyle\lim_{x \to a} f(x) = L$ and $\displaystyle\lim_{x \to a} g(x) = M$ both exist. Then:

1. Constant Multiple Rule

$$\lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x) = c \cdot L$$

You can pull constants out of limits.

Example: When It Works ✓

Problem: If $\lim_{x \to 3} f(x) = 4$, find $\lim_{x \to 3} 5f(x)$.

$$\lim_{x \to 3} 5f(x) = 5 \cdot \lim_{x \to 3} f(x) = 5 \cdot 4 = \boxed{20}$$

Example: When It Fails ✗

Problem: Find $\lim_{x \to 0} \frac{3}{x^2}$.

Why it fails: You can pull out the 3, giving $3 \cdot \lim_{x \to 0} \frac{1}{x^2}$, but $\lim_{x \to 0} \frac{1}{x^2} = \infty$ (doesn't exist as a finite number). The rule requires the inner limit to exist!

2. Sum/Difference Rule

$$\lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) = L \pm M$$

The limit of a sum equals the sum of the limits.

Example: When It Works ✓

Problem: Find $\lim_{x \to 2} (x + x^2)$.

$$\lim_{x \to 2} (x + x^2) = \lim_{x \to 2} x + \lim_{x \to 2} x^2 = 2 + 4 = \boxed{6}$$

Example: When It Fails ✗

Problem: Find $\lim_{x \to \infty} (x - x)$.

Why it fails: You might think $\lim_{x \to \infty} x - \lim_{x \to \infty} x = \infty - \infty$, but $\infty - \infty$ is indeterminate! The rule only works when both limits exist as finite numbers. (The actual answer is 0, but you need to simplify first: $x - x = 0$.)

3. Product Rule

$$\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) = L \cdot M$$

The limit of a product equals the product of the limits.

Example: When It Works ✓

Problem: Find $\lim_{x \to 3} (x \cdot x^2)$.

$$\lim_{x \to 3} (x \cdot x^2) = \lim_{x \to 3} x \cdot \lim_{x \to 3} x^2 = 3 \cdot 9 = \boxed{27}$$

Example: When It Fails ✗

Problem: Find $\lim_{x \to 0} \left(x \cdot \frac{1}{x}\right)$.

Why it fails: You might try $\lim_{x \to 0} x \cdot \lim_{x \to 0} \frac{1}{x} = 0 \cdot \infty$, but $0 \cdot \infty$ is indeterminate! Simplify first: $x \cdot \frac{1}{x} = 1$, so the limit is 1.

4. Quotient Rule

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\displaystyle\lim_{x \to a} f(x)}{\displaystyle\lim_{x \to a} g(x)} = \frac{L}{M} \quad \text{(if } M \neq 0\text{)}$$

The limit of a quotient equals the quotient of the limits (as long as the denominator isn't zero).

Example: When It Works ✓

Problem: Find $\lim_{x \to 4} \frac{x^2}{x}$.

$$\lim_{x \to 4} \frac{x^2}{x} = \frac{\lim_{x \to 4} x^2}{\lim_{x \to 4} x} = \frac{16}{4} = \boxed{4}$$

Example: When It Fails ✗

Problem: Find $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$.

Why it fails: Direct substitution gives $\frac{0}{0}$, which is indeterminate. You can't use the quotient rule when the denominator's limit is 0! Factor first: $\frac{(x-2)(x+2)}{x-2} = x + 2$, so the limit is 4.

5. Power Rule

$$\lim_{x \to a} [f(x)]^n = \left[\lim_{x \to a} f(x)\right]^n = L^n$$

The limit of a power equals the power of the limit.

Example: When It Works ✓

Problem: Find $\lim_{x \to 2} (x + 1)^3$.

$$\lim_{x \to 2} (x + 1)^3 = \left(\lim_{x \to 2} (x + 1)\right)^3 = 3^3 = \boxed{27}$$

Example: When It Fails ✗

Problem: Find $\lim_{x \to 0^+} \left(\frac{1}{x}\right)^2$.

Why it fails: You'd need $\left(\lim_{x \to 0^+} \frac{1}{x}\right)^2 = (\infty)^2$, but $\infty$ is not a number you can raise to a power using this rule. The limit is $\infty$, but you determine this by analyzing the function's behavior, not by applying the power rule.

6. Root Rule

$$\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)} = \sqrt[n]{L}$$

Works when $n$ is odd, or when $n$ is even and $L \geq 0$.

Example: When It Works ✓

Problem: Find $\lim_{x \to 5} \sqrt{x + 4}$.

$$\lim_{x \to 5} \sqrt{x + 4} = \sqrt{\lim_{x \to 5} (x + 4)} = \sqrt{9} = \boxed{3}$$

Example: When It Fails ✗

Problem: Find $\lim_{x \to 3} \sqrt{x - 5}$.

Why it fails: $\lim_{x \to 3} (x - 5) = -2$, and $\sqrt{-2}$ is not a real number! For even roots, you need $L \geq 0$. This limit does not exist in the real numbers.

Special Simple Limits

These are building blocks you'll use constantly:

$$\lim_{x \to a} c = c \quad \text{(constant)}$$

$$\lim_{x \to a} x = a \quad \text{(identity)}$$

$$\lim_{x \to a} x^n = a^n \quad \text{(power)}$$

Direct Substitution

For polynomial and rational functions (where the denominator isn't zero), you can just plug in the value:

$$\lim_{x \to a} p(x) = p(a)$$

This works because polynomials are continuous everywhere!

Example Problems

Example 1: Using Multiple Properties

Problem: Evaluate $\displaystyle\lim_{x \to 2} (3x^2 - 4x + 5)$

Step 1: Use the Sum/Difference Rule to split it up

$$\lim_{x \to 2} (3x^2 - 4x + 5) = \lim_{x \to 2} 3x^2 - \lim_{x \to 2} 4x + \lim_{x \to 2} 5$$

Step 2: Use the Constant Multiple Rule

$$= 3\lim_{x \to 2} x^2 - 4\lim_{x \to 2} x + 5$$

Step 3: Use the basic limits ($\lim x^n = a^n$)

$$= 3(2)^2 - 4(2) + 5$$ $$= 3(4) - 8 + 5$$ $$= 12 - 8 + 5$$ $$\boxed{= 9}$$

Shortcut: Since this is a polynomial, just plug in $x = 2$: $$3(2)^2 - 4(2) + 5 = 12 - 8 + 5 = 9 \checkmark$$

Example 2: Quotient with Direct Substitution

Problem: Evaluate $\displaystyle\lim_{x \to 3} \frac{x^2 + 1}{x - 1}$

Step 1: Check if the denominator equals zero at $x = 3$

$$3 - 1 = 2 \neq 0 \quad \checkmark$$

Step 2: Since the denominator isn't zero, use the Quotient Rule (or just substitute)

$$\lim_{x \to 3} \frac{x^2 + 1}{x - 1} = \frac{3^2 + 1}{3 - 1} = \frac{10}{2} = \boxed{5}$$

Example 3: Using the Root Rule

Problem: Evaluate $\displaystyle\lim_{x \to 4} \sqrt{x^2 + 9}$

Step 1: Apply the Root Rule

$$\lim_{x \to 4} \sqrt{x^2 + 9} = \sqrt{\lim_{x \to 4} (x^2 + 9)}$$

Step 2: Evaluate the inner limit

$$= \sqrt{4^2 + 9} = \sqrt{16 + 9} = \sqrt{25} = \boxed{5}$$

When Direct Substitution FAILS

Direct substitution fails when you get:

$\frac{0}{0}$ — indeterminate form (need algebraic tricks)
$\frac{\text{number}}{0}$ — usually means infinity or DNE

When this happens, you need other techniques:

Factoring and canceling
Rationalizing (multiply by conjugate)
L'Hôpital's Rule (later topic)

Example 4: When You Can't Just Plug In

Problem: Evaluate $\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

Step 1: Try direct substitution

$$\frac{2^2 - 4}{2 - 2} = \frac{0}{0} \quad \text{Indeterminate!}$$

Step 2: Factor and simplify

$$\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2 \quad (\text{for } x \neq 2)$$

Step 3: Now evaluate

$$\lim_{x \to 2} (x + 2) = 2 + 2 = \boxed{4}$$

Formulas & Reference

Constant Multiple Rule

\lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x)

You can pull constants out of limits. Only works if the limit of f(x) exists as a finite number.

Variables:

$c$:: any constant
$\lim_{x \to a} f(x)$:: must exist (finite)

Sum/Difference Rule

\lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)

The limit of a sum (or difference) equals the sum (or difference) of the limits. Only works if both individual limits exist as finite numbers.

Variables:

$\lim_{x \to a} f(x)$:: must exist (finite)
$\lim_{x \to a} g(x)$:: must exist (finite)

Product Rule

\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)

The limit of a product equals the product of the limits. Only works if both individual limits exist as finite numbers.

Variables:

$\lim_{x \to a} f(x)$:: must exist (finite)
$\lim_{x \to a} g(x)$:: must exist (finite)

Quotient Rule

\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}

The limit of a quotient equals the quotient of the limits. Only works if both individual limits exist as finite numbers AND the denominator's limit is not zero.

Variables:

$\lim_{x \to a} f(x)$:: must exist (finite)
$\lim_{x \to a} g(x)$:: must exist (finite) and ≠ 0

Power Rule

\lim_{x \to a} [f(x)]^n = \left[\lim_{x \to a} f(x)\right]^n

The limit of a power equals the power of the limit. Only works if the limit of f(x) exists as a finite number.

Variables:

$\lim_{x \to a} f(x)$:: must exist (finite)
$n$:: any positive integer

Root Rule

\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)}

The limit of a root equals the root of the limit. Only works if the limit of f(x) exists as a finite number. For even roots, the limit must also be non-negative.