Lesson 1.4 · Unit 1 · CED topic 1.5
Determining limits using algebraic properties of limits
Estimating limits from graphs and tables is useful, but it is slow and it is not a proof. Most limits you meet can be found exactly, in a line or two, by using a short list of limit properties. These properties let you take a complicated expression apart, find the limit of each piece, and put the pieces back together.
Learning objective.
- LIM-1.D. Determine the limits of functions using limit theorems.
Essential knowledge.
- LIM-1.D.1. One-sided limits can be determined analytically or graphically.
- LIM-1.D.2. Limits of sums, differences, products, quotients, and composite functions can be found using limit theorems.
The limit properties
Suppose \(\displaystyle\lim_{x\to c}f(x)=A\) and \(\displaystyle\lim_{x\to c}g(x)=B\), where \(A\) and \(B\) are real numbers, and let \(k\) be a constant. Then:
| Property | Statement |
|---|---|
| Constant | \(\displaystyle\lim_{x\to c}k=k\) |
| Identity | \(\displaystyle\lim_{x\to c}x=c\) |
| Constant multiple | \(\displaystyle\lim_{x\to c}\,[k\,f(x)]=kA\) |
| Sum and difference | \(\displaystyle\lim_{x\to c}\,[f(x)\pm g(x)]=A\pm B\) |
| Product | \(\displaystyle\lim_{x\to c}\,[f(x)\,g(x)]=AB\) |
| Quotient | \(\displaystyle\lim_{x\to c}\dfrac{f(x)}{g(x)}=\dfrac{A}{B}\), provided \(B\ne0\) |
| Power | \(\displaystyle\lim_{x\to c}\,[f(x)]^n=A^n\) (and roots, when they are defined) |
| Composite | \(\displaystyle\lim_{x\to c}F(f(x))=F(A)\) if \(F\) is continuous at \(A\) |
In words: limits of sums, differences, products, quotients, and composite functions can be found using limit theorems, as long as the limits of the pieces exist (and, for a quotient, the bottom limit is not zero). The same properties hold for one-sided limits, where \(x\to c^-\) or \(x\to c^+\) throughout.
Direct substitution
Apply the constant and identity properties repeatedly and you get a powerful shortcut. For any polynomial \(p\), \(\displaystyle\lim_{x\to c}p(x)=p(c)\). For a rational function \(\dfrac{p(x)}{q(x)}\), \(\displaystyle\lim_{x\to c}\dfrac{p(x)}{q(x)}=\dfrac{p(c)}{q(c)}\) as long as \(q(c)\ne0\). The same is true for the other familiar functions (roots, exponentials, logarithms, sine, cosine) wherever they are defined. This is direct substitution, and it is the first thing to try on any limit. When it yields a real number, you are done. If it gives \(\frac{0}{0}\), the property of the quotient does not apply and you need another method (Lesson 1.6). If it gives a nonzero number divided by zero, the limit is unbounded (Lesson 1.14).
Find \(\displaystyle\lim_{x\to2}(3x^2-2x+1)\) and justify each step.
\[\lim_{x\to2}(3x^2-2x+1)=3\Big(\lim_{x\to2}x\Big)^2-2\lim_{x\to2}x+\lim_{x\to2}1=3(2)^2-2(2)+1=12-4+1=9.\]
The constant multiple, power, difference, and sum properties allowed us to evaluate each piece separately. This agrees with direct substitution, \(p(2)=9\).
Find \(\displaystyle\lim_{x\to2}\frac{\sqrt{x^2+5}}{x+1}\).
The numerator's limit is \(\sqrt{4+5}=3\) and the denominator's limit is \(2+1=3\neq0\). By the quotient property, the limit is \(\dfrac{3}{3}=1\).
Suppose \(\displaystyle\lim_{x\to a}f(x)=3\) and \(\displaystyle\lim_{x\to a}g(x)=-2\). Find \(\displaystyle\lim_{x\to a}\dfrac{2f(x)-[g(x)]^2}{f(x)+g(x)}\).
Numerator: \(2(3)-(-2)^2=6-4=2\). Denominator: \(3+(-2)=1\ne0\). Quotient: \(\dfrac{2}{1}=2\).
You never need a formula for \(f\) and \(g\). The properties work from the limit values alone.
Let \(f(x)=\begin{cases}x^2-1,&x\lt2\\ 7-x,&x\ge2.\end{cases}\) Find \(\displaystyle\lim_{x\to2}f(x)\).
Left. For \(x\lt2\), \(f(x)=x^2-1\), so \(\displaystyle\lim_{x\to2^-}f(x)=2^2-1=3\).
Right. For \(x\ge2\) (and near 2 on the right), \(f(x)=7-x\), so \(\displaystyle\lim_{x\to2^+}f(x)=7-2=5\).
The one-sided limits differ, so \(\displaystyle\lim_{x\to2}f(x)\) does not exist. Here \(f(2)=5\) is defined, but that does not repair the disagreement.
Composite functions
The composite property says you may "push the limit inside" a function that is continuous at the limit value. For example, \(\displaystyle\lim_{x\to0}\cos(x^2+\pi)=\cos\!\Big(\lim_{x\to0}(x^2+\pi)\Big)=\cos\pi=-1\). Be careful: if the outside function is not continuous at the inner limit, pushing the limit inside can give a wrong answer. Continuity gets its own treatment in Lessons 1.10 through 1.13.
What to write on the exam
Show the substitution and the value: \(\displaystyle\lim_{x\to3}\frac{x^2-1}{x+2}=\frac{3^2-1}{3+2}=\frac{8}{5}\). If an algebraic property requires a condition (nonzero denominator, existence of the pieces), check it. A statement such as "the limit of the quotient is the quotient of the limits" is only valid when the bottom limit is not zero.
- Try direct substitution first for polynomials, rational functions (denominator nonzero), and the familiar functions on their domains.
- The sum, difference, constant multiple, product, quotient, and power properties require the limits of the pieces to exist; the quotient also requires a nonzero bottom limit.
- For a piecewise function, find each one-sided limit with the piece that applies on that side, then compare.
- \(\frac{0}{0}\) from substitution means "not finished," not "does not exist."
- Evaluate \(\displaystyle\lim_{x\to-1}(x^3-4x+7)\).
Show answer
By direct substitution, \((-1)^3-4(-1)+7=-1+4+7=10\).
- Evaluate \(\displaystyle\lim_{x\to3}\dfrac{x^2-1}{x+2}\).
Show answer
The denominator is \(3+2=5\ne0\), so by substitution the limit is \(\dfrac{9-1}{5}=\dfrac{8}{5}\).
- Evaluate \(\displaystyle\lim_{x\to4}\sqrt{x+5}\cdot(x-1)\).
Show answer
\(\sqrt{9}\cdot3=3\cdot3=9\).
- Suppose \(\displaystyle\lim_{x\to a}f(x)=-4\) and \(\displaystyle\lim_{x\to a}g(x)=5\). Find \(\displaystyle\lim_{x\to a}[f(x)g(x)+2f(x)]\).
Show answer
By the product, constant multiple, and sum properties: \((-4)(5)+2(-4)=-20-8=-28\).
- Let \(g(x)=\begin{cases}2x+1,&x\le3\\ x^2-2,&x\gt3.\end{cases}\) Find \(\displaystyle\lim_{x\to3}g(x)\).
Show answer
Left: \(2(3)+1=7\). Right: \(3^2-2=7\). The one-sided limits agree, so \(\displaystyle\lim_{x\to3}g(x)=7\).
- Explain why the quotient property cannot be used to evaluate \(\displaystyle\lim_{x\to1}\dfrac{x-1}{x^2-1}\), and then evaluate the limit another way.
Show answer
The limit of the denominator is \(1^2-1=0\), and the quotient property requires a nonzero bottom limit (substitution gives \(\frac{0}{0}\)). For \(x\ne1\), \(\dfrac{x-1}{x^2-1}=\dfrac{x-1}{(x-1)(x+1)}=\dfrac{1}{x+1}\), whose limit as \(x\to1\) is \(\dfrac{1}{2}\).