The limit laws
Taking limits apart
If f(x)\to 3 and g(x)\to -1 as x\to 1, what does f(x)+g(x) approach? It seems obvious. When does that kind of reasoning break down?
The limit laws say you may take a limit apart, term by term or factor by factor, provided each piece has a limit and you never divide by a limit of 0. Build a combination below and compare what the laws predict with what the function actually does.
Indigo dashes show f, ochre dashes show g, and the solid curve is the combination. All limits are taken as x → 1.
The limit laws
Suppose \lim_{x\to a}f(x) and \lim_{x\to a}g(x) both exist, and c is a constant. Then:
| Law | Statement |
|---|---|
| Sum and difference | \lim\,[f(x)\pm g(x)]=\lim f(x)\pm\lim g(x) |
| Constant multiple | \lim\,c\,f(x)=c\lim f(x) |
| Product | \lim\,f(x)g(x)=\lim f(x)\cdot\lim g(x) |
| Quotient | \lim\dfrac{f(x)}{g(x)}=\dfrac{\lim f(x)}{\lim g(x)}\ \text{ if } \lim g(x)\ne 0 |
| Power and root | \lim\,[f(x)]^n=\big[\lim f(x)\big]^n,\quad \lim\sqrt[n]{f(x)}=\sqrt[n]{\lim f(x)} |
For a root with n even, the limit of f(x) must be positive. Every law has the same condition: the pieces must have limits. The laws say nothing when a piece has no limit. In that case the combination might still have one. Try the first button in the graph: \sin\tfrac{1}{x-1} has no limit, yet adding its negative gives exactly 0.
Direct substitution
Applying the laws again and again shows that if p is a polynomial, then \lim_{x\to a}p(x)=p(a). The same holds for a rational function p/q as long as q(a)\ne 0. That is why substituting is always the first thing to try. Why does it fail when q(a)=0?
Solve it step by step
Use the limit laws to evaluate each limit, or decide that they cannot be used. Answers can be fractions such as -3/2. The score counts steps you answer correctly on the first try.
Common mistakes
- Using the quotient law when the bottom approaches 0
If \lim g(x)=0, the quotient law does not apply. You need another method, such as factoring, or the conclusion that no finite limit exists.
- Concluding too much from a failed law
If \lim f(x) does not exist, that does not prove \lim\,[f(x)+g(x)] fails to exist. The law is silent, not negative.
- Splitting when a piece has no limit
Writing \lim x\sin\tfrac1x=\lim x\cdot\lim\sin\tfrac1x is invalid, because the second limit does not exist. The squeeze theorem handles this one.