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Calculus AB · Unit 1 · 1.5
AP Calculus · Limits and Continuity
Determining Limits Using Algebraic Properties
Split the expression, take each limit, reassemble — but only when every piece exists.
key idea$\lim(f\pm g)=\lim f\pm\lim g,\quad \lim(fg)=\lim f\cdot\lim g$
The rule
The limit laws let you take a complicated expression apart, take the limit of each simple piece, and put the answer back together. They all carry the same precondition: every individual limit you split into must itself exist.
The limit laws
\[\lim_{x\to c}\bigl[f(x)\pm g(x)\bigr]=L\pm M\]
If $\lim_{x\to c}f=L$ and $\lim_{x\to c}g=M$ (both finite), then sums, differences, products, constant multiples and quotients all distribute. For the quotient, $M\neq 0$ is required.
The precondition is the whole rule
Every law begins “if both limits exist.” Split a limit whose pieces do not individually exist and you can produce nonsense — see the last worked example.
Pattern recognition
| Law | Statement | Condition |
|---|
| Sum / difference | $\lim(f\pm g)=L\pm M$ | both exist |
| Constant multiple | $\lim(kf)=kL$ | $L$ exists |
| Product | $\lim(fg)=L\cdot M$ | both exist |
| Quotient | $\lim\dfrac{f}{g}=\dfrac{L}{M}$ | $M\neq0$ |
| Power | $\lim f^{n}=L^{n}$ | $L$ exists |
| Root | $\lim\sqrt[n]{f}=\sqrt[n]{L}$ | $L\ge 0$ for even $n$ |
| Composition | $\lim f(g(x))=f(M)$ | $f$ continuous at $M$ |
The right-hand column is what gets tested. A law quoted without its condition is an incomplete justification.
working order
- Substitute first — if it gives a number, the laws already did their work.
- Split only into pieces whose limits you can name.
- A $\tfrac{0}{0}$ form means the quotient law does not apply — go to 1.6.
- $\lim$ of a constant is that constant; $\lim_{x\to c}x=c$. Everything builds from these two.
Worked examples
EX 1
Take it apart, put it back
$3x^{2}-5x+2$Evaluate $\lim\limits_{x\to2}\left(3x^{2}-5x+2\right)$ using the laws, naming each one.
1
Split the sum
By the sum/difference law: $\lim 3x^{2}-\lim 5x+\lim 2$.
2
Pull out constants
Constant multiple law: $3\lim x^{2}-5\lim x+2$.
3
Use the two base limits
Power law on the first: $3(2)^{2}-5(2)+2=12-10+2=4$.
4
Note what you proved
Every piece existed, so the split was legal. For a polynomial this always works — which is exactly why polynomials are continuous everywhere.
EX 2
Quotient law with its condition
$\dfrac{x^{2}+1}{x-3}$Evaluate $\lim\limits_{x\to1}\dfrac{x^{2}+1}{x-3}$.
1
Check the denominator first
$\lim_{x\to1}(x-3)=-2\neq0$, so the quotient law applies. Check this before splitting, not after.
2
Split
$\dfrac{\lim(x^{2}+1)}{\lim(x-3)}$.
3
Evaluate each
Numerator $\to 2$; denominator $\to -2$.
4
Combine
$\dfrac{2}{-2}=-1$.
EX 3
From a table of limits
structuralGiven $\lim\limits_{x\to4}f(x)=6$ and $\lim\limits_{x\to4}g(x)=-2$, find $\lim\limits_{x\to4}\dfrac{2f(x)}{f(x)+g(x)}$.
1
Check the denominator
$\lim(f+g)=6+(-2)=4\neq0$. The quotient law is available.
2
Apply sum and constant multiple
Numerator: $2\lim f=12$. Denominator: $\lim f+\lim g=4$.
3
Combine
$\dfrac{12}{4}=3$.
4
Why no formulas were needed
The laws are structural — they work on the limit values alone, which is why this form appears so often on the exam.
EX 4
Splitting an illegal split
the trapA student writes $\lim\limits_{x\to0}\left(\dfrac1x-\dfrac1x\right)=\lim\dfrac1x-\lim\dfrac1x=0$. Evaluate the reasoning.
1
Check the answer
The expression is identically $0$ for every $x\neq0$, so the limit really is $0$. The answer is right.
2
Check the reasoning
The difference law requires both limits to exist. $\lim_{x\to0}\tfrac1x$ does not exist, so the split was never legal.
3
Show why it matters
The same illegal move on $\lim_{x\to0}\left(\tfrac1{x^{2}}-\tfrac1{x^{2}}\right)$ would “give” $\infty-\infty$, which is meaningless.
4
Do it correctly
Simplify first: the expression is $0$ for all $x\neq0$, so the limit is $0$ by the constant rule. Right answer, wrong method loses the justification mark.
Watch out
The quotient law dies when $M=0$
A zero denominator limit means the law simply does not apply. It does not mean the limit is undefined — rewrite the expression, or analyse each side separately.
“Both limits exist” is not decoration
Example 4 gets the right number by an invalid route. On a justification question that is a lost mark.
$\infty$ is not a value you may substitute
You cannot write $\lim f=\infty$ and then use $f$ in the sum law. Unbounded behaviour needs its own analysis, not the limit laws.
Composition needs continuity
$\lim f(g(x))=f\left(\lim g\right)$ only when $f$ is continuous at that inner limit. It is not a free move.