Choose a useful first step
Three quotients at x = 1
The buttons show three quotients. A is \(\dfrac{x+3}{x^2+1}\), B is \(\dfrac{x+3}{x-1}\) and C is \(\dfrac{x^2+2x-3}{x-1}\). The orange dot marks the curve at the input x, and the readouts show the top, the bottom and the quotient.
On A, drag x toward 1. The top approaches 4 and the bottom approaches 2, so the quotient approaches 2.
On B, the top approaches 4 and the bottom approaches 0, so the quotient becomes very large. It is positive on the right of 1 and negative on the left, so B has no limit.
On C, the top and the bottom both approach 0. The curve has an open circle, called a ring, at x = 1 because C has no value there, but the dot still approaches height 4.
Read the top and the bottom
The curve is now any quotient \(\dfrac{N(x)}{D(x)}\), and the faint line marks the input \(c\). The orange bars show the top \(\cy{N(c)}\) and the bottom \(\cy{D(c)}\), and the sliders change them.
If the bottom bar is not 0, the curve has a height at \(c\), and the limit is the top divided by the bottom. If only the bottom bar is 0, the curve rises on one side of \(c\) and falls on the other, so there is no limit.
If both bars are 0, a ring appears at \(c\) because the quotient has no value there, but the curve may still approach a height. In every case, the values at \(c\) decide what happens, so you substitute \(c\) first.
Notation
Substitute \(c\) into the top and into the bottom.
\[\cy{N(c)}\quad\text{and}\quad \cy{D(c)}\]If the bottom is not 0, the limit is the value of the quotient, because the quotient has a height at \(c\).
\[\lim_{x\to c}\frac{N(x)}{D(x)}=\cy{\frac{N(c)}{D(c)}}\]If the bottom is 0 and the top is not, the quotient grows without bound near \(c\), so the limit does not exist.
\[\frac{\cy{N(c)}}{\cy{0}}\ \text{with}\ N(c)\neq 0\]If both are 0, the form \(\dfrac00\) is not a number, so you rewrite the expression.
\[\frac{\cy{0}}{\cy{0}}\]For polynomials, factor and cancel. For a square root, multiply the top and the bottom by the conjugate, which has the opposite sign between its terms. The conjugate of \(\sqrt{x}-3\) is \(\sqrt{x}+3\).
The rewrite is valid because a limit ignores the value at \(c\), and the new expression has the same values at every other input.
When it fails
When both bars are 0, the quotient has no value at \(c\), so the dot disappears at \(c\) and substitution cannot give the limit.
A missing value does not mean a missing limit, because the curve on both sides of the ring is close to the same height.
Example 1: substitute, then factor
Find \(\displaystyle\lim_{x\to 2}\frac{x^2+x-6}{x^2-4}\).
Substitute \(x = 2\). The top is \(2^2+2-6 = 0\) and the bottom is \(2^2-4 = 0\), so the form is \(\dfrac00\).
\[\frac{0}{0}\]Both are polynomials, so factor them. Each is 0 at \(x=2\), so each contains \(x-2\).
\[\frac{(x+3)(x-2)}{(x+2)(x-2)}\]A limit never uses \(x=2\) itself, so \(x-2\) is not 0 and you can cancel it. The orange dashed curve shows the new expression.
\[\frac{x+3}{x+2}\]The new bottom at \(x=2\) is 4, so substitute.
\[\frac{2+3}{2+2}=\cy{\frac{5}{4}}\]
Example 2: a square root
Find \(\displaystyle\lim_{x\to 9}\frac{\sqrt{x}-3}{x-9}\).
Substitute \(x=9\). The top is \(\sqrt{9}-3=0\) and the bottom is \(9-9=0\), so the form is \(\dfrac00\).
\[\frac{0}{0}\]A tempting wrong move: The top is 0, so the quotient is 0 and the limit is 0.
\[\lim_{x\to 9}\frac{\sqrt{x}-3}{x-9}=\cr{0}\]The form \(\dfrac00\) is not 0, because the bottom is 0 too. Near \(x=9\), the curve stays near height \(\tfrac16\), well above the red line at 0.
The expression has a square root, so multiply the top and the bottom by the conjugate \(\sqrt{x}+3\). This multiplies the quotient by 1, so its value does not change.
\[\begin{gathered}\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{(x-9)(\sqrt{x}+3)}\\ =\frac{x-9}{(x-9)(\sqrt{x}+3)}\end{gathered}\]Cancel \(x-9\), because \(x\) is not 9 in a limit. The new bottom at \(x=9\) is \(\sqrt{9}+3=6\), so substitute.
\[\frac{1}{\sqrt{x}+3}\to \cy{\frac{1}{6}}\]
Practice
Find \(\displaystyle\lim_{x\to -1}\frac{x^2-x-2}{x^2+3x+2}\).
Substitute first. If the form is \(\dfrac00\), the graph has a ring at \(x=-1\), and the limit is the height around it.
Solution
Substitute \(x=-1\). The top is \(1+1-2=0\) and the bottom is \(1-3+2=0\), so the form is \(\dfrac00\).
\[\frac{0}{0}\]Factor both polynomials. Each one contains the factor \(x+1\).
\[\frac{(x-2)(x+1)}{(x+2)(x+1)}\]Cancel \(x+1\). The new bottom at \(x=-1\) is 1, so substitute.
\[\frac{x-2}{x+2}\to\frac{-3}{1}=\cy{-3}\]
Questions
The graph shows \(q(x)=\dfrac{x^2+2x-15}{x^2-2x-3}\), with a ring at \(x=3\) and a vertical asymptote at \(x=-1\).
As \(x\) approaches 3, which value does \(q(x)\) approach?
- 2
- 0
- The limit does not exist, because \(q(3)\) is undefined
- 3
Answer
2
As \(x\) approaches \(-1\), which statement is correct?
- The limit does not exist, because the top is \(-16\) and the bottom is 0
- The limit is \(-16\)
- The limit is 2, as at x = 3
- The limit is 0, because the bottom is 0
Answer
The limit does not exist, because the top is \(-16\) and the bottom is 0
Find \(\displaystyle\lim_{x\to -3}\frac{x^2-9}{x^2+5x+6}\).
- 6
- 0
- \(-6\)
- The limit does not exist
Answer
6
Summary
You substitute the input into the top and the bottom first, because the two values tell you which method fits.
A bottom that is not 0 gives the limit, and a top that is not 0 over a bottom of 0 gives no limit.
For \(\dfrac00\), you factor or use the conjugate, which keeps every value except the one at the input, so the limit stays the same.
This is one of four skills in topic 1.7 of Which & Why: AP Calculus.