Lesson 1.5 · Unit 1 · CED topic 1.6
Determining limits using algebraic manipulation
Direct substitution is the first move, but it sometimes returns \(\frac{0}{0}\), as it did for \(\dfrac{x^2-4}{x-2}\) at \(x=2\). That form is called indeterminate: it says nothing about the limit by itself, because the limit could be any number, or not exist. The cure is algebra. The graph of the function is the same as the graph of a simpler function except at the one troublesome point, so we rearrange the expression into an equivalent form that substitution can handle. Limits depend only on values near \(c\), and for \(x\ne c\) the two forms are identical.
Learning objective.
- LIM-1.E. Determine the limits of functions using equivalent expressions for the function or the squeeze theorem.
Essential knowledge.
- LIM-1.E.1. It may be necessary or helpful to rearrange expressions into equivalent forms before evaluating limits.
Technique 1: factor and cancel
When substitution into a rational function gives \(\frac{0}{0}\), the numerator and denominator share a factor \((x-c)\). Factor, divide it out (legitimate because \(x\ne c\) in a limit), and substitute again.
Find \(\displaystyle\lim_{x\to3}\frac{x^2-9}{x^2-5x+6}\).
Direct substitution gives \(\frac{0}{0}\). Factor: \(\dfrac{(x-3)(x+3)}{(x-3)(x-2)}=\dfrac{x+3}{x-2}\) for \(x\ne3\). Now substitute: \(\dfrac{3+3}{3-2}=6\).
\[\lim_{x\to3}\frac{x^2-9}{x^2-5x+6}=\lim_{x\to3}\frac{x+3}{x-2}=6.\]
Technique 2: multiply by the conjugate
When the \(\frac{0}{0}\) involves a square root, multiply the numerator and denominator by the conjugate of the radical expression, so that the difference of squares removes the root. Keep the other factor unmultiplied so that you can cancel.
Find \(\displaystyle\lim_{h\to0}\frac{\sqrt{9+h}-3}{h}\).
Substitution gives \(\frac{0}{0}\). Multiply by \(\dfrac{\sqrt{9+h}+3}{\sqrt{9+h}+3}\):
\[\frac{\sqrt{9+h}-3}{h}\cdot\frac{\sqrt{9+h}+3}{\sqrt{9+h}+3}=\frac{(9+h)-9}{h\left(\sqrt{9+h}+3\right)}=\frac{1}{\sqrt{9+h}+3}.\]
Now \(h\to0\) gives \(\dfrac{1}{\sqrt9+3}=\dfrac{1}{6}\). This is the derivative of \(\sqrt{x}\) at \(x=9\) in disguise, as Unit 2 will show.
Technique 3: combine fractions
When the expression is a fraction of fractions (a complex fraction), put the numerator over a common denominator, simplify, and cancel.
Find \(\displaystyle\lim_{x\to0}\frac{\dfrac{1}{x+2}-\dfrac12}{x}\).
The numerator is \(\dfrac{2-(x+2)}{2(x+2)}=\dfrac{-x}{2(x+2)}\). Dividing by \(x\) gives \(\dfrac{-x}{2(x+2)\,x}=\dfrac{-1}{2(x+2)}\) for \(x\ne0\). Then
\[\lim_{x\to0}\frac{-1}{2(x+2)}=\frac{-1}{2\cdot2}=-\frac14.\]
Technique 4: trigonometric limits and identities
Two limits do most of the work with trigonometric functions. With \(x\) in radians,
\[\lim_{x\to0}\frac{\sin x}{x}=1 \qquad\text{and}\qquad \lim_{x\to0}\frac{1-\cos x}{x}=0.\]
The first was suggested by the table in Lesson 1.4 and is justified with the squeeze theorem in Lesson 1.8; the second follows from it. Use them together with alternate forms of trigonometric functions, such as \(\tan x=\dfrac{\sin x}{\cos x}\) or the identity \(1-\cos x=\dfrac{\sin^2x}{1+\cos x}\), to rewrite an expression in a form where these limits apply. A frequent move is to scale the argument: \(\dfrac{\sin(kx)}{x}=k\cdot\dfrac{\sin(kx)}{kx}\to k\cdot1\).
Find (a) \(\displaystyle\lim_{x\to0}\frac{\sin5x}{3x}\) and (b) \(\displaystyle\lim_{x\to0}\frac{\tan x}{x}\).
(a) \(\dfrac{\sin5x}{3x}=\dfrac{5}{3}\cdot\dfrac{\sin5x}{5x}\). As \(x\to0\), \(5x\to0\) too, so \(\dfrac{\sin5x}{5x}\to1\), and the limit is \(\dfrac53\).
(b) \(\dfrac{\tan x}{x}=\dfrac{\sin x}{x}\cdot\dfrac{1}{\cos x}\to1\cdot\dfrac{1}{\cos0}=1\cdot1=1\).
Choosing the technique
Look at the form of the expression. A rational function with a shared factor: factor. A radical difference: conjugate. A fraction inside a fraction: common denominator. A trigonometric function near \(0\): use \(\frac{\sin x}{x}\to1\), or rewrite with identities. Lesson 1.7 collects these choices into one decision guide.
- \(\frac{0}{0}\) is indeterminate: it calls for rearranging the expression, not for answering "does not exist."
- Factor and cancel the shared \((x-c)\); multiply by the conjugate for radicals; use a common denominator for complex fractions.
- Know \(\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1\) and \(\displaystyle\lim_{x\to0}\frac{1-\cos x}{x}=0\) (radians), and the scaling trick \(\dfrac{\sin kx}{x}\to k\).
- After simplifying, write the equivalent expression and then substitute; do not substitute into the original and write \(\frac{0}{0}=\) a number.
- Evaluate \(\displaystyle\lim_{x\to2}\dfrac{x^2-4}{x^2-3x+2}\).
Show answer
Factor: \(\dfrac{(x-2)(x+2)}{(x-2)(x-1)}=\dfrac{x+2}{x-1}\). Substituting \(x=2\) gives \(\dfrac{4}{1}=4\).
- Evaluate \(\displaystyle\lim_{x\to0}\dfrac{\sqrt{x+16}-4}{x}\).
Show answer
Multiply by the conjugate: \(\dfrac{x}{x(\sqrt{x+16}+4)}=\dfrac{1}{\sqrt{x+16}+4}\to\dfrac{1}{8}\).
- Evaluate \(\displaystyle\lim_{x\to1}\dfrac{x-1}{\sqrt{x}-1}\).
Show answer
Multiply by \(\dfrac{\sqrt x+1}{\sqrt x+1}\): \(\dfrac{(x-1)(\sqrt x+1)}{x-1}=\sqrt x+1\to2\). (Or note \(x-1=(\sqrt x-1)(\sqrt x+1)\).)
- Evaluate \(\displaystyle\lim_{x\to0}\dfrac{\frac{1}{x+3}-\frac13}{x}\).
Show answer
The numerator is \(\dfrac{3-(x+3)}{3(x+3)}=\dfrac{-x}{3(x+3)}\), so the quotient is \(\dfrac{-1}{3(x+3)}\to\dfrac{-1}{9}\).
- Evaluate \(\displaystyle\lim_{x\to0}\dfrac{\sin7x}{2x}\).
Show answer
\(\dfrac{\sin7x}{2x}=\dfrac72\cdot\dfrac{\sin7x}{7x}\to\dfrac72\cdot1=\dfrac72\).
- Evaluate \(\displaystyle\lim_{x\to0}\dfrac{\tan3x}{x}\).
Show answer
\(\dfrac{\tan3x}{x}=\dfrac{\sin3x}{x\cos3x}=3\cdot\dfrac{\sin3x}{3x}\cdot\dfrac{1}{\cos3x}\to3\cdot1\cdot1=3\).