AP Calculus AB Evaluating Limits Algebraically Lesson (Free PDF)

AP Calculus AB math lesson: Evaluating Limits Algebraically – Maths PDF

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This limit algebra lesson for AP Calculus AB covers evaluating limits algebraically (Common Core AP CALC AB 1.5, AP CALC AB 1.6, AP CALC AB 1.7, AP CALC AB 1.8), with a free PDF to print.

To evaluate a limit algebraically, first try direct substitution. If you get the indeterminate form 0/0, factor, rationalize with a conjugate, or simplify, cancel the common factor, and substitute again. The limit laws let you combine known limits, and the Squeeze Theorem handles functions trapped between two others.

AP Calculus AB · Common Core State Standards: AP CALC AB 1.5, AP CALC AB 1.6, AP CALC AB 1.7, AP CALC AB 1.8

Key facts in this limit algebra lesson

Key facts

Direct substitution works when the function is defined and continuous at the target.
The form 0/0 means “simplify more”, not “the limit is 0” and not “does not exist”.
Multiply by the conjugate to remove a square root from a limit.
lim sin(x)/x = 1 and lim (1 – cos x)/x = 0 as x approaches 0 (radians).

What are the limit laws?

If \(\lim_{x\to a} f(x)=L\) and \(\lim_{x\to a} g(x)=K\) both exist, limits can be combined with arithmetic.

Limit laws

\(\lim [f(x)\pm g(x)]=L\pm K\)
\(\lim [c\cdot f(x)]=cL\)
\(\lim [f(x)\cdot g(x)]=L\cdot K\)
\(\lim \frac{f(x)}{g(x)}=\frac{L}{K}\) when \(K\neq 0\)
\(\lim [f(x)]^{n}=L^{n}\)

For polynomials and most functions you meet in class, direct substitution works: \(\lim_{x\to a} f(x)=f(a)\) whenever \(f(a)\) is defined.

Try direct substitution first, then simplify only if you get 0/0.

Use this quick guide to choose a method before you start any algebra.

What you see after substituting	What to do
A real number	That number is the limit.
\(\frac{0}{0}\) with polynomials	Factor, cancel the common factor, substitute again.
\(\frac{0}{0}\) with a square root	Multiply by the conjugate, simplify, substitute again.
A sine or cosine over \(x\)	Use the special trig limits.
A bounded oscillating factor	Try the Squeeze Theorem.

What if substitution gives 0/0?

The form \(\frac{0}{0}\) is indeterminate: the limit may still exist. Factor the numerator and denominator, cancel the common factor, then substitute again.

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

The graph is the line y = x − 1 with a hole at (−2, −3).

How do you handle square roots?

Multiply the top and bottom by the conjugate to remove the radical. For \(\lim_{x\to 9}\frac{\sqrt{x}-3}{x-9}\), multiply by \(\frac{\sqrt{x}+3}{\sqrt{x}+3}\) to get \(\frac{x-9}{(x-9)(\sqrt{x}+3)}=\frac{1}{\sqrt{x}+3}\), so the limit is \(\frac{1}{6}\).

The graph of (√x − 3)/(x − 9) has a hole at (9, 1/6).

Which trig limits must you know?

Special trigonometric limits

\(\lim_{x\to 0}\frac{\sin x}{x}=1\) and \(\lim_{x\to 0}\frac{1-\cos x}{x}=0\). The angle must be in radians. Rewrite \(\frac{\sin 3x}{x}\) as \(3\cdot\frac{\sin 3x}{3x}\) to get \(3\).

The graph of sin x / x approaches 1 at x = 0.

What is the Squeeze Theorem?

If \(g(x)\le f(x)\le h(x)\) near \(a\) and \(\lim g(x)=\lim h(x)=L\), then \(\lim f(x)=L\). Because \(-x^{2}\le x^{2}\sin\frac{1}{x}\le x^{2}\), the middle limit at \(0\) is \(0\).

What is a clean written method?

Substitute the target value and look at the form of the result.
If the form is \(\frac{0}{0}\), pick factoring, a conjugate or a special limit.
Simplify completely and cancel the common factor, writing “for \(x\neq a\)” when it helps.
Substitute again and state the limit as a number.

Limit algebra lesson: questions and answers

What is the first step in evaluating a limit?

Substitute the target value directly. If the result is a real number, that is the limit. If you get 0/0, simplify the expression and try again.

Does 0/0 mean the limit does not exist?

No. The form 0/0 is indeterminate, so the limit could be any number or may not exist. Factor or use a conjugate to find out.

How do you use a conjugate to find a limit?

Multiply the numerator and denominator by the conjugate of the radical expression. The product of conjugates removes the square root and often creates a factor you can cancel.

Why is the limit of sin(x)/x equal to 1?

For small angles in radians, sin x is very close to x, so the ratio is very close to 1. The Squeeze Theorem proves it rigorously.

When can I use the Squeeze Theorem?

Use it when a hard function is trapped between two simpler functions that approach the same limit at the target. Then the hard function has that limit too.

How to use this limit algebra lesson

Read the definitions first, then follow the worked method line by line and redo the example on paper without looking. When every step makes sense, move on to the worksheet, then check yourself with the quiz.

Download the limit algebra lesson as a PDF with the button below and keep it with your notes.

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