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lim
\( x \to a \)

Algebraic Properties
of Limits

Master the fundamental limit laws that let you break complex limits into simpler pieces: sums, products, quotients, and powers.

AP Calculus BC

What Is a Limit?

A limit describes the value a function \( f(x) \) approaches as the input \( x \) gets closer and closer to a target value \( a \). We write this as:

The key insight: we care about what happens near \( a \), not necessarily at \( a \). The function doesn't even need to be defined at \( a \) for the limit to exist.

Visualizing Limit Behavior

x y L a ← → approaches L

Why Do Limit Laws Matter?

Without limit laws, you'd need to prove every limit from the \( \varepsilon \)-\( \delta \) definition. The algebraic properties of limits let you decompose complex expressions into manageable parts.

Simplify
Break a complicated limit into sums, products, and quotients of simpler limits.
Evaluate
Compute each piece separately, then combine results using the laws.
Justify
Provide rigorous reasoning on the AP exam by citing specific limit properties.

The Core Limit Laws

Given that \( \lim_{x \to a} f(x) = L \) and \( \lim_{x \to a} g(x) = M \), these properties hold:

1
Sum Law: \( \lim_{x \to a} [f(x) + g(x)] = L + M \)
2
Difference Law: \( \lim_{x \to a} [f(x) - g(x)] = L - M \)
3
Constant Multiple Law: \( \lim_{x \to a} [c \cdot f(x)] = c \cdot L \)
4
Product Law: \( \lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M \)
5
Quotient Law: \( \lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{L}{M} \), provided \( M \neq 0 \)

Power and Root Laws

The limit laws extend naturally to powers and roots, completing our algebraic toolkit:

6
Power Law: \( \lim_{x \to a} [f(x)]^n = L^n \) for any positive integer \( n \)
7
Root Law: \( \lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L} \), provided \( L \geq 0 \) when \( n \) is even

Direct Substitution: The First Strategy

When \( f \) is a polynomial or rational function (and the denominator is nonzero at \( a \)), the limit laws combine to give us the simplest possible approach:

This works because polynomials are built from sums, products, and constant multiples of \( x \), and each limit law applies at every step. When direct substitution gives \( \frac{0}{0} \), we need algebraic manipulation first.

Direct Substitution vs. Indeterminate Forms

Direct Substitution Works Indeterminate Form
What happens Plugging in \( a \) gives a finite value Plugging in \( a \) gives \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)
Next step Done: that value is the limit Simplify first (factor, rationalize, etc.)
Example \( \lim_{x \to 2}(3x+1) = 7 \) \( \lim_{x \to 1}\frac{x^2-1}{x-1} \)
Limit law used Sum, Constant Multiple Quotient Law fails: \( M=0 \)

Step-by-Step Strategy

1
Try direct substitution. Plug \( a \) into \( f(x) \). If you get a finite number, that's the limit.
2
Check for indeterminate form. If substitution gives \( \frac{0}{0} \), algebraic simplification is needed.
3
Simplify the expression. Factor, cancel common factors, rationalize, or expand as needed.
4
Apply limit laws. Use sum, product, quotient, power, and root laws to evaluate the simplified expression.
5
Substitute again. After simplification, try direct substitution on the new expression to get the final answer.

Example 1: Sum and Power Laws

Evaluate \( \displaystyle\lim_{x \to 3} \left(2x^2 - 5x + 4\right) \).

1
Apply the Sum/Difference Law: \( \lim_{x\to 3}(2x^2) - \lim_{x\to 3}(5x) + \lim_{x\to 3}(4) \)
2
Apply the Constant Multiple Law: \( 2\cdot\lim_{x\to 3}(x^2) - 5\cdot\lim_{x\to 3}(x) + 4 \)
3
Apply the Power Law and substitute: \( 2(3)^2 - 5(3) + 4 = 18 - 15 + 4 \)

Example 2: Quotient Law with Factoring

Evaluate \( \displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2} \).

1
Try direct substitution: \( \frac{2^2-4}{2-2} = \frac{0}{0} \). Indeterminate form.
2
Factor the numerator: \( \frac{(x-2)(x+2)}{x-2} \)
3
Cancel the common factor \( (x-2) \): this is valid since \( x \neq 2 \) when taking the limit.
4
Evaluate: \( \lim_{x\to 2}(x+2) = 2 + 2 \)

Example 3: Rationalizing a Root

Evaluate \( \displaystyle\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} \).

1
Direct substitution gives \( \frac{\sqrt{4}-2}{0} = \frac{0}{0} \). Indeterminate.
2
Multiply by conjugate: \( \frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} = \frac{(x+4)-4}{x(\sqrt{x+4}+2)} \)
3
Simplify: \( \frac{x}{x(\sqrt{x+4}+2)} = \frac{1}{\sqrt{x+4}+2} \)
4
Now substitute: \( \frac{1}{\sqrt{0+4}+2} = \frac{1}{2+2} \)

Visualizing the Factoring Example

x y 2 4 hole at (2, 4) \( f(x) = \frac{x^2-4}{x-2} \) \( = x + 2,\; x \neq 2 \)

Combining Multiple Laws

Evaluate \( \displaystyle\lim_{x \to 1} \frac{3x^2 + \sqrt{x}}{x + 5} \).

1
Check denominator: \( 1 + 5 = 6 \neq 0 \). Quotient Law applies.
2
Numerator (Sum Law): \( \lim_{x\to 1}(3x^2) + \lim_{x\to 1}(\sqrt{x}) \)
3
Constant Multiple + Power + Root: \( 3(1)^2 + \sqrt{1} = 3 + 1 = 4 \)
4
Quotient Law: \( \frac{4}{6} = \frac{2}{3} \)

Common Mistakes to Avoid

✕ Applying Quotient Law When \( M = 0 \)
The Quotient Law requires \( \lim g(x) \neq 0 \). If the denominator approaches 0, you must simplify first.
✕ Declaring "DNE" Too Quickly
Getting \( \frac{0}{0} \) doesn't mean the limit doesn't exist. It means you need to factor, rationalize, or use another technique.
✕ Confusing \( f(a) \) with \( \lim_{x\to a} f(x) \)
The function value at \( a \) and the limit as \( x \to a \) are different concepts. The limit can exist even when \( f(a) \) is undefined.

Limit Laws Summary

Law Formula Condition
Sum / Difference \( L \pm M \) Both limits exist
Constant Multiple \( c \cdot L \) \( L \) exists
Product \( L \cdot M \) Both limits exist
Quotient \( L / M \) Both exist, \( M \neq 0 \)
Power \( L^n \) \( L \) exists, \( n \in \mathbb{Z}^+ \)
Root \( \sqrt[n]{L} \) \( L \geq 0 \) if \( n \) even

Question 1 of 5

Given \( \lim_{x\to 3} f(x) = 5 \) and \( \lim_{x\to 3} g(x) = -2 \), what is \( \lim_{x\to 3}[4f(x) - g(x)] \)?

Question 2 of 5

What is \( \displaystyle\lim_{x\to 4} \frac{x^2 - 16}{x - 4} \)?

Question 3 of 5

Which limit law cannot be applied when \( \lim_{x\to a} g(x) = 0 \)?

Question 4 of 5

Evaluate \( \displaystyle\lim_{x\to 9} \frac{\sqrt{x} - 3}{x - 9} \).

Question 5 of 5

If \( \lim_{x\to 2} f(x) = 3 \), what is \( \lim_{x\to 2} \left[f(x)\right]^3 \)?