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lim

Determining Limits Using
Algebraic Manipulation

Master the essential techniques for evaluating limits when direct substitution yields indeterminate forms like \(\frac{0}{0}\).

AP Calculus BC

What Is a Limit?

A limit describes the value a function \(f(x)\) approaches as the input \(x\) gets arbitrarily close to some number \(c\). We write:

$$\lim_{x \to c} f(x) = L$$

This means: as \(x \to c\), the outputs \(f(x) \to L\). The function does not need to equal \(L\) at \(x = c\): the limit cares about approach, not arrival.

The Indeterminate Form Problem

When direct substitution gives \(\frac{0}{0}\), the limit is indeterminate: it could be any real number (or fail to exist). Consider:

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

Plugging in \(x = 2\) yields \(\frac{0}{0}\). But if we factor the numerator:

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

Visualizing the Hole

x y 0 1 2 3 4 5 2 4 6
hole at \(x = 2\)
\(y = x + 2\)
\(L = 4\)

Core Techniques for \(\frac{0}{0}\)

Factoring
Factor numerator and denominator, then cancel the common factor causing \(\frac{0}{0}\).
Conjugate
Multiply by the conjugate to eliminate radicals and simplify the expression.
Trig Identities
Apply identities like \(\sin^2\theta + \cos^2\theta = 1\) or known limits to simplify.

The Limit Laws Foundation

Every algebraic manipulation is justified by the Limit Laws. If \(\lim_{x \to c} f(x)\) and \(\lim_{x \to c} g(x)\) both exist:

1
Sum/Difference: \(\lim_{x \to c}[f(x) \pm g(x)] = \lim f(x) \pm \lim g(x)\)
2
Product: \(\lim_{x \to c}[f(x) \cdot g(x)] = \lim f(x) \cdot \lim g(x)\)
3
Quotient: \(\lim_{x \to c}\frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}\) provided \(\lim g(x) \neq 0\)
4
Power: \(\lim_{x \to c}[f(x)]^n = [\lim f(x)]^n\)

Indeterminate vs. Determinate Forms

Form Type What to Do
\(\frac{0}{0}\) Indeterminate Simplify algebraically, then re-substitute
\(\frac{k}{0}\), \(k \neq 0\) Determinate Limit is \(\pm\infty\) or DNE (check signs)
\(\frac{0}{k}\), \(k \neq 0\) Determinate Limit is \(0\)
\(\frac{k}{m}\), both \(\neq 0\) Determinate Limit is \(\frac{k}{m}\) by direct substitution

Step-by-Step Approach

1
Try direct substitution. Plug \(x = c\) into \(f(x)\). If you get a real number, you are done.
2
Identify the form. If you get \(\frac{0}{0}\), proceed to algebraic manipulation. If \(\frac{k}{0}\), investigate for \(\pm\infty\).
3
Simplify. Factor, multiply by conjugate, use trig identities, or combine fractions to cancel the troublesome factor.
4
Substitute again. After simplification, plug in \(x = c\) to evaluate the limit.
5
Verify. Check your answer by testing values near \(c\) or by graphing the original function.

Technique 1: Factoring

Evaluate: \(\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3}\)

1
Direct substitution: \(\frac{3^2 - 9}{3 - 3} = \frac{0}{0}\) : indeterminate.
2
Factor: \(\frac{(x-3)(x+3)}{x-3}\). Cancel the common factor \((x-3)\).
3
Simplify: \(= x + 3\) for \(x \neq 3\).
4
Substitute: \(\lim_{x \to 3}(x+3) = \) \(6\).

Technique 2: Conjugate Multiplication

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

1
Direct substitution: \(\frac{\sqrt{4} - 2}{0} = \frac{0}{0}\) : indeterminate.
2
Multiply by conjugate: \(\frac{\sqrt{x+4}-2}{x} \cdot \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}\)
3
Simplify numerator: \((\sqrt{x+4})^2 - 4 = x+4-4 = x\). So: \(\frac{x}{x(\sqrt{x+4}+2)}\).
4
Cancel and substitute: \(\frac{1}{\sqrt{x+4}+2} \to \frac{1}{\sqrt{4}+2} = \) \(\frac{1}{4}\).

Technique 3: Trigonometric Limits

Two foundational trig limits you must memorize:

\(\displaystyle\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1\)
The Squeeze Theorem establishes this fundamental result.
\(\displaystyle\lim_{\theta \to 0}\frac{1 - \cos\theta}{\theta} = 0\)
Derived from the first limit using the identity \(1-\cos\theta = \frac{\sin^2\theta}{1+\cos\theta}\).

Example: Trig Limit in Action

Evaluate: \(\displaystyle\lim_{x \to 0}\frac{\sin(3x)}{\sin(5x)}\)

1
Direct substitution: \(\frac{\sin(0)}{\sin(0)} = \frac{0}{0}\) : indeterminate.
2
Rewrite strategically: Multiply and divide to create the standard form: $$\frac{\sin(3x)}{\sin(5x)} = \frac{\sin(3x)}{3x}\cdot\frac{5x}{\sin(5x)}\cdot\frac{3x}{5x}$$
3
Apply known limits: As \(x \to 0\): \(\frac{\sin(3x)}{3x} \to 1\), \(\frac{5x}{\sin(5x)} \to 1\), and \(\frac{3x}{5x} = \frac{3}{5}\).
4
Result: \(1 \cdot 1 \cdot \frac{3}{5} = \) \(\frac{3}{5}\).

Example: Complex Fraction

Evaluate: \(\displaystyle\lim_{x \to 1}\frac{\frac{1}{x} - 1}{x - 1}\)

1
Direct substitution: \(\frac{\frac{1}{1} - 1}{1 - 1} = \frac{0}{0}\) : indeterminate.
2
Simplify the numerator: \(\frac{1}{x} - 1 = \frac{1 - x}{x}\). So the expression becomes \(\frac{\frac{1-x}{x}}{x-1}\).
3
Rewrite as multiplication: \(\frac{1-x}{x} \cdot \frac{1}{x-1} = \frac{-(x-1)}{x(x-1)}\).
4
Cancel and substitute: \(\frac{-1}{x} \to \frac{-1}{1} = \) \(-1\).

Example: Cubic Factoring

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

1
Direct substitution: \(\frac{(-1)^3 + 1}{-1+1} = \frac{0}{0}\) : indeterminate.
2
Factor (sum of cubes): \(x^3 + 1 = (x+1)(x^2 - x + 1)\).
3
Cancel: \(\frac{(x+1)(x^2 - x + 1)}{x+1} = x^2 - x + 1\).
4
Substitute: \((-1)^2 - (-1) + 1 = 1 + 1 + 1 = \) \(3\).

Common Mistakes

✕ "The limit is 0/0"
\(\frac{0}{0}\) is not a number and not a limit. It signals that more work is needed: factor, conjugate, or simplify first.
✕ Canceling without noting domain
Writing \(\frac{x-3}{x-3} = 1\) is only valid for \(x \neq 3\). Always state the restriction when canceling.
✕ Wrong conjugate pairing
For \(\sqrt{a} - b\), multiply by \(\frac{\sqrt{a}+b}{\sqrt{a}+b}\), not \(\frac{\sqrt{a}-b}{\sqrt{a}-b}\). Multiplying by itself does nothing useful.

Technique Summary

Technique When to Use Key Move
Factoring Polynomial numerator and/or denominator Factor out and cancel common \((x-c)\) term
Conjugate Expression contains \(\sqrt{\phantom{x}}\) Multiply top and bottom by conjugate
Trig Identities Expression has \(\sin, \cos, \tan\) Rewrite to match \(\frac{\sin\theta}{\theta} \to 1\)
Complex Fraction Fraction within a fraction Combine into single fraction, then simplify

Question 1 of 5

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

Question 2 of 5

Evaluate \(\displaystyle\lim_{x \to 0}\frac{\sqrt{x+9} - 3}{x}\).

Question 3 of 5

Evaluate \(\displaystyle\lim_{x \to 0}\frac{\sin(7x)}{x}\).

Question 4 of 5

Evaluate \(\displaystyle\lim_{x \to -2}\frac{x^3 + 8}{x + 2}\).

Question 5 of 5

Which algebraic technique is most appropriate for evaluating \(\displaystyle\lim_{x \to 4}\frac{\sqrt{x}-2}{x-4}\)?