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Calculus BC · Unit 1 · 1.6
AP Calculus · Limits and Continuity
Determining Limits Using Algebraic Manipulation
A $0/0$ form means a factor cancels. Find it, rewrite, and substitute again.
key idea$\tfrac00\ \Rightarrow\ \text{find an equivalent expression, then substitute again}$
The rule
When substitution gives $\tfrac00$, the function has a removable problem at that point. Rewrite it as an expression that agrees with the original everywhere except at $c$ — and since a limit never looks at $x=c$, the rewrite has the same limit.
Equivalent expressions
If $f(x)=g(x)$ for all $x\neq c$, then $\lim_{x\to c}f=\lim_{x\to c}g$. This is what makes cancelling legal: you are not dividing by zero, because $x$ never equals $c$ inside a limit.
Why $\tfrac00$ is the signal
A $\tfrac00$ form means numerator and denominator share a factor that vanishes at $c$. Find it and cancel it. A $\tfrac{k}{0}$ form shares nothing — no manipulation exists.
Pattern recognition
| When the expression has… | Technique | What appears after |
|---|
| a factorable polynomial | factor and cancel | substitute again |
| a square root in a sum | multiply by the conjugate | the root disappears from one side |
| a fraction inside a fraction | clear the inner denominators | a single simple quotient |
| $\sin$ or $\cos$ near $0$ | reshape to $\tfrac{\sin u}{u}$ | a known limit of $1$ |
| absolute values | split by sign | two one-sided limits |
| a difference of powers | $a^{n}-b^{n}$ factoring | a cancelling factor |
Each row is triggered by what you can see in the expression, so the skill is recognition before computation.
Worked examples
EX 1
Factor and cancel
$\dfrac{x^{2}-x-6}{x-3}$Evaluate $\lim\limits_{x\to3}\dfrac{x^{2}-x-6}{x-3}$.
1
Substitute
$\tfrac{9-3-6}{0}=\tfrac00$ — indeterminate, so a common factor exists.
2
Factor the numerator
$x^{2}-x-6=(x-3)(x+2)$.
3
Cancel legally
For $x\neq3$, $\dfrac{(x-3)(x+2)}{x-3}=x+2$. The restriction $x\neq3$ is exactly what a limit already assumes.
4
Substitute again
$\lim_{x\to3}(x+2)=5$.
EX 2
Conjugate
$\dfrac{\sqrt{x}-2}{x-4}$Evaluate $\lim\limits_{x\to4}\dfrac{\sqrt{x}-2}{x-4}$.
1
Substitute
$\tfrac{2-2}{0}=\tfrac00$.
2
Multiply by the conjugate
$\dfrac{\sqrt x-2}{x-4}\cdot\dfrac{\sqrt x+2}{\sqrt x+2}=\dfrac{x-4}{(x-4)\left(\sqrt x+2\right)}$.
3
Cancel
$=\dfrac{1}{\sqrt x+2}$ for $x\neq4$.
4
Substitute again
$\dfrac{1}{2+2}=\dfrac14$.
5
Note the shortcut
$x-4=\left(\sqrt x-2\right)\left(\sqrt x+2\right)$ — recognising that difference of squares gets you there in one step.
EX 3
Compound fraction
$\dfrac{\frac1x-\frac13}{x-3}$Evaluate $\lim\limits_{x\to3}\dfrac{\frac1x-\frac13}{x-3}$.
1
Substitute
$\tfrac{\frac13-\frac13}{0}=\tfrac00$.
2
Combine the inner fractions
$\dfrac1x-\dfrac13=\dfrac{3-x}{3x}$.
3
Divide by $x-3$
$\dfrac{3-x}{3x}\cdot\dfrac{1}{x-3}=\dfrac{-(x-3)}{3x(x-3)}=\dfrac{-1}{3x}$.
4
Substitute again
$\dfrac{-1}{9}$.
5
The sign trap
$3-x=-(x-3)$. Dropping that minus is the commonest error in this pattern.
EX 4
Trig near zero
$\dfrac{1-\cos x}{x^{2}}$Evaluate $\lim\limits_{x\to0}\dfrac{1-\cos x}{x^{2}}$.
1
Substitute
$\tfrac{1-1}{0}=\tfrac00$.
2
Multiply by the conjugate
$\dfrac{1-\cos x}{x^{2}}\cdot\dfrac{1+\cos x}{1+\cos x}=\dfrac{1-\cos^{2}x}{x^{2}(1+\cos x)}$.
3
Use the identity
$1-\cos^{2}x=\sin^{2}x$, giving $\dfrac{\sin^{2}x}{x^{2}(1+\cos x)}=\left(\dfrac{\sin x}{x}\right)^{2}\cdot\dfrac{1}{1+\cos x}$.
4
Apply the known limits
$\left(\dfrac{\sin x}{x}\right)^{2}\to1$ and $\dfrac{1}{1+\cos x}\to\dfrac12$.
5
Combine
The limit is $\dfrac12$.
Watch out
Manipulation only helps a $\tfrac00$ form
With $\tfrac{k}{0}$ there is no shared factor, so no rewrite exists. Check the form before reaching for algebra.
Cancelling is not dividing by zero
It is legal precisely because $x\neq c$ throughout a limit. Say so if a question asks you to justify the step.
The rewrite is a different function
$\dfrac{x^{2}-x-6}{x-3}$ and $x+2$ agree everywhere except $x=3$, where the first is undefined. Same limit, different functions.
Sign slips when flipping a factor
$a-b=-(b-a)$. Reversing a factor to cancel it costs a minus sign that is easy to lose — Example 3.