Wemazi

Topic 1.6

Recognize an indeterminate form

Builds on recognizing when a limit law is inconclusive. 6 minutes.

Three quotients at x = 2

The buttons show three quotients near \(x=2\). A is \(\tfrac{x^2+1}{x+3}\), B is \(\tfrac{x+3}{x-2}\) and C is \(\tfrac{x^2-x-2}{x-2}\). For each one, drag the orange dot toward 2.

On A the dot approaches height 1. On B the dot rises without limit on the right of 2 and falls without limit on the left. On C the dot approaches height 3, and the curve has a hole at \(x=2\).

Put the input at 2 and read the top and the bottom. A gives 5 and 5, B gives 5 and 0, and C gives 0 and 0.

The numbers for A and B decide the limit. The numbers for C do not, and the next scene shows why.

What the top and the bottom tell you

The graph has no numbers. The horizontal position of the orange dot is \(\cy{T}\), the value of the top at the input. Its height is \(\cy{B}\), the value of the bottom.

The red strip along the horizontal axis is where \(\cy{B}\) is 0. Above or below it, \(\cy{B}\) is not 0, so the limit of the quotient is \(\cy{T}\) over \(\cy{B}\).

In the red strip, away from the center, \(\cy{B}\) is 0 and \(\cy{T}\) is not. The quotient is very large near the input, so there is no limit.

In the dashed square at the center, \(\cy{T}\) and \(\cy{B}\) are both 0. This form, 0 over 0, is called indeterminate because the form alone does not decide the limit.

Notation

  1. For a quotient \(\frac{f(x)}{g(x)}\), substitute the input \(c\) into the top and the bottom. If \(f\) and \(g\) are continuous at \(c\), their limits are these two values.

    \[\cy{T}=f(c),\qquad \cy{B}=g(c)\]

  2. If \(B\) is not 0, the limit of the quotient is \(T\) over \(B\). On graph A, this is 5 over 5.

    \[\lim_{x\to c}\frac{f(x)}{g(x)}=\frac{\cy{T}}{\cy{B}}\]

  3. If \(B\) is 0 and \(T\) is not, the quotient has no limit, because the top stays near \(T\) while the bottom shrinks toward 0.

  4. If \(T\) and \(B\) are both 0, the form is indeterminate. The limit may be any number, or it may not exist.

    \[\frac{\cy{0}}{\cy{0}}\]

When it fails

The curve is the quotient \(\dfrac{k(x-2)}{x-2}\) for a number \(k\) that you choose. Whatever \(k\) is, substituting 2 gives 0 over 0.

At every input other than 2, the quotient equals \(k\), so the limit is \(k\). The form is 0 over 0 for every \(k\), so it cannot tell you the limit.

Example 1: A form that needs more work

Substitute 4 into \(\dfrac{x^2-16}{x^2-5x+4}\). What does the result tell you about the limit as \(x\) approaches 4?

  1. Substitute 4 into the top.

    \[4^2-16=0\]

  2. Substitute 4 into the bottom.

    \[4^2-5(4)+4=0\]

  3. Both are 0, so the form is 0 over 0 and does not decide the limit. The dots on both sides of the hole approach the orange line at height \(\tfrac83\), so the limit exists.

    \[\lim_{x\to 4}\frac{x^2-16}{x^2-5x+4}=\cy{\frac83}\]

Example 2: A zero that is not 0 over 0

Substitute \(-2\) into \(\dfrac{x^2-1}{x+2}\). What does the result tell you about the limit as \(x\) approaches \(-2\)?

  1. Substitute −2 into the top and into the bottom.

    \[(-2)^2-1=3,\qquad -2+2=0\]

  2. A tempting wrong move: The bottom is 0, so this is 0 over 0. Look for a common factor to cancel.

    \[\frac{3}{0}\ \text{is indeterminate}\]

    The top is 3, so the form is 3 over 0, not 0 over 0. The graph has no hole at x = −2, only a dashed line where the curve rises and falls without limit.

  3. The bottom approaches 0 while the top approaches 3, so the quotient becomes very large. It is positive on the right of −2 and negative on the left, so the limit does not exist.

Practice

Substitute \(x=3\) into \(\dfrac{x^2-2x-3}{x^2-9}\) and into \(\dfrac{x^2+2x-3}{x^2-9}\). Which quotient is indeterminate? What do you know about the other?

Find the value of the top and the value of the bottom at 3 in each quotient. Only a top and a bottom that are both 0 give 0 over 0.

Solution

  1. In the first quotient, the top and the bottom are both 0, so it is indeterminate.

    \[\begin{gathered}3^2-2(3)-3=0\\ 3^2-9=0\end{gathered}\]

  2. In the second quotient, the top is 12 and the bottom is 0. The form is not 0 over 0, so there is no limit at 3.

    \[3^2+2(3)-3=12\]

Questions

The graph shows two quotients, \(P\) and \(Q\). An open ring marks a missing point, and the dashed vertical line marks \(x=-3\).

  1. Substituting \(x=1\) into \(P\) gives 0 over 0. What does the graph show about the limit of \(P\) as \(x\) approaches 1?

    • The limit is 3
    • The limit is 0
    • The limit does not exist
    Answer

    The limit is 3

  2. Substituting \(x=-3\) into \(Q\) gives 4 over 0. Which statement is correct?

    • The form is not 0 over 0, and the limit does not exist
    • The form is 0 over 0, so the limit may exist
    • The limit is 4
    Answer

    The form is not 0 over 0, and the limit does not exist

  3. Which limit is an indeterminate form 0 over 0 when \(x\) approaches 2?

    • \(\dfrac{x^3-8}{x-2}\)
    • \(\dfrac{x^3+8}{x-2}\)
    • \(\dfrac{x^3-8}{x+2}\)
    • \(\dfrac{x^2+4}{x+2}\)
    Answer

    \(\dfrac{x^3-8}{x-2}\)

Summary

Substituting the input into the top and the bottom sorts a quotient into one of three cases.

Only 0 over 0 leaves the limit open, because the other two cases give the limit or show that there is none.

When you see 0 over 0, rewrite the quotient, because the form alone does not give the limit.

This is one of four skills in topic 1.6 of Which & Why: AP Calculus.