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lim x → a f

Algebraic Properties
of Limits

Learn how to break apart complex limits into simpler pieces using the powerful algebraic limit laws: sums, products, quotients, and more.

AP Calculus AB

What Is a Limit?

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

The limit can exist even if \( f(a) \) is undefined: it only depends on the behaviour near \( a \), not at \( a \) itself.

Why Learn Limit Laws?

Most limits you encounter in AP Calculus are combinations of simpler functions: sums, differences, products, quotients, and powers. The algebraic properties of limits let you split a complex expression into pieces you can evaluate individually.

Simplify
Break one hard limit into several easy ones.
Justify
Provide rigorous reasoning on the AP exam.
Build
These laws underpin derivatives and integrals.

Visual Intuition

If two functions each approach a value as \( x \to a \), their sum approaches the sum of those values. The graph below shows \( f(x) \) and \( g(x) \) converging, and their sum \( f(x)+g(x) \) converging to \( L_1 + L_2 \).

x y a \(L_1\) \(L_2\) \(L_1\!+\!L_2\) f(x) g(x) f(x)+g(x)

The Core Limit Laws

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

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

Power & Root Laws

Two more laws extend the toolkit to exponents and radicals:

Power Law
\(\displaystyle\lim_{x \to a}[f(x)]^n = L^n\) for any positive integer \(n\).
Root Law
\(\displaystyle\lim_{x \to a}\sqrt[n]{f(x)} = \sqrt[n]{L}\) when defined (if \(n\) is even, require \(L \geq 0\)).

Direct Substitution Principle

For polynomials and rational functions (where the denominator is nonzero), the limit laws combine to give the simplest possible rule:

This works because each building-block limit (\(\lim_{x\to a} c = c\) and \(\lim_{x\to a} x = a\)) exists, and the limit laws let you combine them freely.

When Can't You Use Direct Substitution?

Scenario Direct Sub Works Direct Sub Fails
Result of plugging in \(a\) Finite number \(\frac{0}{0}\) or \(\frac{k}{0}\)
What to do Done: that's the limit Simplify first (factor, rationalize, etc.)
Example \(\lim_{x\to 3}(x^2+1) = 10\) \(\lim_{x\to 2}\frac{x^2-4}{x-2}\)
Limit law status All laws apply directly Quotient law fails (\(M=0\))

Step-by-Step Strategy

1
Try direct substitution. Plug \(a\) into every \(x\).
2
Check the result. If you get a finite number, you're done.
3
If you get \(\frac{0}{0}\): Factor, cancel, rationalize, or use another algebraic technique to simplify.
4
Substitute again in the simplified expression.
5
Cite the laws. On AP free-response, name each law you invoke (Sum, Product, etc.).

Example 1: Sum & Constant Multiple

Evaluate \(\displaystyle\lim_{x \to 2}\bigl(3x^2 + 5x - 7\bigr)\).

1
Direct substitution: Plug in \(x = 2\).
2
\(3(2)^2 + 5(2) - 7 = 12 + 10 - 7 = 15\).
3
No \(\frac{0}{0}\): the result is finite.

Example 2: Quotient Law

Evaluate \(\displaystyle\lim_{x \to 4}\frac{x^2 - 1}{x + 3}\).

1
Substitute \(x = 4\): numerator \(= 16 - 1 = 15\), denominator \(= 4 + 3 = 7\).
2
Denominator limit is \(7 \neq 0\), so the Quotient Law applies.
3
\(\displaystyle\frac{15}{7}\) is the final answer.

Example 3: Handling \(\frac{0}{0}\)

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

1
Substitute \(x = 2\): \(\frac{4-4}{2-2} = \frac{0}{0}\). Indeterminate: must simplify.
2
Factor: \(x^2 - 4 = (x-2)(x+2)\), so \(\frac{(x-2)(x+2)}{x-2} = x + 2\) for \(x \neq 2\).
3
Substitute again: \(\lim_{x\to 2}(x+2) = 4\).

Example 4: Root Law in Action

Evaluate \(\displaystyle\lim_{x \to 9}\sqrt{x + 7}\).

1
Inner limit: \(\lim_{x\to 9}(x+7) = 16\) by the Sum Law.
2
\(16 \geq 0\), so the Root Law applies.
3
\(\sqrt{16} = 4\).

Combining Multiple Laws

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

1
Numerator: \(\lim_{x\to 1}(2x+1)=3\), then Power Law gives \(3^3=27\).
2
Denominator: \(\lim_{x\to 1}(x+3)=4\), then Root Law gives \(\sqrt{4}=2\).
3
Quotient Law: Denominator limit is \(2 \neq 0\), so \(\frac{27}{2}\).

Common Mistakes

✕ Dividing by zero
Applying the Quotient Law when the denominator limit is 0. Always check \(M \neq 0\) first.
✕ Assuming \(\frac{0}{0}\) means DNE
\(\frac{0}{0}\) is indeterminate, not nonexistent. Simplify and try again.
✕ Forgetting the root condition
The Root Law for even \(n\) requires \(L \geq 0\). If \(L < 0\), the real root doesn't exist.

Summary of Limit Laws

Law Statement Condition
Sum \(\lim[f+g]=L+M\) Both exist
Difference \(\lim[f-g]=L-M\) Both exist
Constant Multiple \(\lim[cf]=cL\) Limit exists
Product \(\lim[f \cdot g]=L \cdot M\) Both exist
Quotient \(\lim\frac{f}{g}=\frac{L}{M}\) \(M \neq 0\)
Power \(\lim[f]^n = L^n\) Limit exists
Root \(\lim\sqrt[n]{f}=\sqrt[n]{L}\) \(L \geq 0\) if \(n\) even

Question 1 of 5

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

Question 2 of 5

Evaluate \(\displaystyle\lim_{x \to -1}(4x^3 - 2x + 6)\).

Question 3 of 5

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

Question 4 of 5

Evaluate \(\displaystyle\lim_{x \to 5}\frac{x^2 - 25}{x - 5}\).

Question 5 of 5

If \(\lim_{x\to 2}f(x)=9\), what is \(\displaystyle\lim_{x\to 2}\sqrt{f(x)}\)?