Limit laws
Limit Laws
Limit Laws is a topic in Limits and Continuity in the California Calculus Standards. It is aligned to Standard 1.1, which requires students to prove and use the theorems for evaluating the limits of sums, products, quotients, and compositions of functions.
The limit laws let you evaluate a complicated limit by breaking it into pieces: the limit of a sum, difference, product, quotient, power, or root is built from the separate limits, provided each one exists.
Theory
The limit laws let you evaluate a complicated limit by breaking it into simpler pieces. If \(\lim_{x\to a} f\) and \(\lim_{x\to a} g\) both exist, the limit of a sum, difference, product, quotient, power, or root is built from the separate limits — with one caution for the quotient law.
The limit laws say that if \(\lim_{x\to a} f(x)\) and \(\lim_{x\to a} g(x)\) both exist, then limits distribute over the basic operations. In words:
- Sum / difference: the limit of \(f\pm g\) is the sum or difference of the limits.
- Constant multiple: a constant factor pulls outside the limit.
- Product: the limit of \(f\cdot g\) is the product of the limits.
- Quotient: the limit of \(\dfrac{f}{g}\) is the quotient of the limits — provided the bottom limit is not \(0\).
- Power / root: the limit moves inside a power or root.
With \(\lim_{x\to a} f=F\) and \(\lim_{x\to a} g=G\):
How to use the limit laws
- Break up the expression using the sum, product, and constant-multiple laws.
- Substitute \(x=a\) into each simple piece.
- For a quotient, check the bottom first. If \(\lim g\neq 0\), divide the limits. If \(\lim g=0\), do not use the quotient law — factor and cancel, then try again.
Split with the difference and constant-multiple laws, then substitute.
| \(\lim_{x\to 3}(2x^2-5x)\) | \(=\) | \(2\lim_{x\to 3}x^2-5\lim_{x\to 3}x\) |
| \(=\) | \(2\cdot 9-5\cdot 3\) | |
| \(=\) | \(18-15=3\) |
The bottom limit is \(1^2+1=2\neq 0\), so the quotient law applies.
| \(\lim_{x\to 1}\dfrac{x+4}{x^2+1}\) | \(=\) | \(\dfrac{\lim_{x\to 1}(x+4)}{\lim_{x\to 1}(x^2+1)}\) |
| \(=\) | \(\dfrac{5}{2}\) |
The root law lets the limit move inside the square root.
| \(\lim_{x\to 2}\sqrt{x^2+5}\) | \(=\) | \(\sqrt{\lim_{x\to 2}(x^2+5)}\) |
| \(=\) | \(\sqrt{9}=3\) |
The bottom limit is \(0\), so the quotient law does not apply directly. Factor and cancel first.
| \(\dfrac{x-2}{x^2-4}\) | \(=\) | \(\dfrac{x-2}{(x-2)(x+2)}=\dfrac{1}{x+2}\) |
| \(\lim_{x\to 2}\dfrac{1}{x+2}\) | \(=\) | \(\dfrac{1}{4}\) |
Common pitfalls
Frequently asked questions
What are the limit laws?
Rules that say the limit of a sum, difference, product, quotient, power, or root equals the same operation applied to the separate limits — as long as those limits exist (and the bottom of a quotient is not zero).
When does the quotient law not work?
When the limit of the denominator is \(0\). Then you cannot divide the limits; factor and cancel the fraction first, then take the limit.
How are the limit laws related to direct substitution?
For polynomials, roots, and rational functions with a nonzero denominator, applying the laws gives the same result as substituting \(x=a\). The laws are the reason substitution works.
Can I use the laws when a piece goes to infinity?
No. The laws require each separate limit to be a finite number. If a piece heads to \(\pm\infty\), handle the expression another way (for example, simplify or compare degrees).