Wemazi

Topic 1.14

Determine the sign on each side

Builds on finding candidate vertical asymptotes. 6 minutes.

Two sides, two signs

The curve shows \(f(x) = \dfrac{4 - x}{x + 1}\), beside the dashed line \(x = -1\). The curve is green above the x-axis, where the outputs are positive, and red below it, where they are negative.

Drag the slider to move the orange point toward \(-1\) from the right. At \(x = -0.9\) the top is 4.9, the bottom is 0.1, and the height is 49.

Press the second button to move the point to the left of \(-1\). At \(x = -1.1\) the top is 5.1, the bottom is \(-0.1\), and the height is \(-51\).

The top stays positive, but the bottom changes sign at −1, so the curve climbs on the right and falls on the left.

The sign of a large quotient

Near the dashed line \(x = c\), the top has one fixed sign. The bottom approaches 0, so the quotient grows large, but the sign of the bottom can depend on the side.

The sign of the quotient is the sign of the top times the sign of the bottom. A positive product gives a green part that climbs, and a negative product gives a red part that falls.

Press the first toggle. The top becomes negative, so both parts reverse their colors.

Press the second toggle. A square is positive on both sides, so both parts take the sign of the top.

Notation

  1. Write the bottom as \((x - c)^n\), the factor that is 0 at \(c\), times the rest, \(B(x)\). If \(N(c)\) and \(B(c)\) are not 0, the rest of the fraction keeps the sign \(s\) of \(\dfrac{N(c)}{B(c)}\) near \(c\).

    \[\begin{gathered} f(x) = \frac{N(x)}{(x - c)^n\,B(x)} \\ s = \text{sign of } \frac{N(c)}{B(c)} \end{gathered}\]

  2. Find the sign of \((x - c)^n\) on each side. Right of \(c\), the factor \(x - c\) is small and positive. Left of \(c\), it is small and negative. An odd power keeps these signs, and an even power is positive on both sides.

  3. On each side, multiply \(s\) by the sign of \((x - c)^n\). A positive product gives \(\infty\), and a negative product gives \(-\infty\).

    \[\lim_{x \to c^+} f(x) = \cg{+\infty} \ \text{ or } \ \cr{-\infty}\]

  4. If both sides give the same infinity, write one two-sided limit. If the sides differ, the two-sided limit does not exist. The symbol \(\infty\) describes outputs that grow without bound, and it is not a number.

When it fails

If the top is also 0 at \(c\), the quotient need not grow large, so this method does not apply. With the toggle on, the asymptote is gone and an open ring marks a hole at \(c\).

Cancel the shared factor first. Then use the sign method on the fraction that remains.

Example 1: a square in the bottom

Find \(\displaystyle\lim_{x \to 2} \frac{x + 5}{(x - 2)^2}\).

  1. The top at \(x = 2\) is \(2 + 5 = 7\), which is positive.

  2. The bottom \((x - 2)^2\) is a square, so it is positive on both sides of 2 while it approaches 0.

  3. A positive number divided by a small positive number is large and positive, so both parts of the curve are green.

    \[\begin{gathered} \lim_{x \to 2^-} f(x) = \cg{\infty} \\ \lim_{x \to 2^+} f(x) = \cg{\infty} \end{gathered}\]

  4. The two sides agree, so the limit is \(\infty\). As a check, the heights at \(x = 1.99\) and \(x = 2.01\) are 69,900 and 70,100.

Example 2: the sides differ

Find the one-sided limits of \(f(x) = \dfrac{x - 7}{x - 2}\) at \(x = 2\).

  1. The top at \(x = 2\) is \(2 - 7 = -5\), which is negative.

  2. A tempting wrong move: The top is negative, so the function goes to −∞ on both sides.

    \[\lim_{x \to 2} f(x) = -\infty\]

    The bottom \(x - 2\) changes sign at 2. At \(x = 1.9\) the height is \(\tfrac{-5.1}{-0.1} = 51\), which is positive, so the left part is green. The dashed red curve shows the fall that this move claims.

  3. Right of 2, the bottom is small and positive, so the quotient is large and negative. Left of 2, the bottom is small and negative, so the quotient is large and positive.

    \[\begin{gathered} \lim_{x \to 2^+} f(x) = \cr{-\infty} \\ \lim_{x \to 2^-} f(x) = \cg{+\infty} \end{gathered}\]

  4. The two sides differ, so the two-sided limit does not exist. The line \(x = 2\) is still a vertical asymptote.

Practice

Find the one-sided limits of \(p(x) = \dfrac{6}{(x + 1)(x - 3)}\) at \(x = -1\).

Find the sign of \(x - 3\) near \(x = -1\). Then find the sign of \(x + 1\) on each side, and compare with the colors.

Solution

  1. Near \(x = -1\), the factor \(x - 3\) is close to \(-4\), which is negative. The top, 6, is positive, so the rest of the fraction is negative.

  2. Right of \(-1\), \(x + 1\) is small and positive, so the quotient is large and negative. Left of \(-1\), \(x + 1\) is small and negative, so the quotient is large and positive.

    \[\begin{gathered} \lim_{x \to -1^+} p(x) = \cr{-\infty} \\ \lim_{x \to -1^-} p(x) = \cg{+\infty} \end{gathered}\]

Questions

The graph shows \(u(x) = \dfrac{4}{(x + 1)^2 (x - 2)}\). Green parts lie above the x-axis, and red parts lie below.

  1. What is \(\displaystyle\lim_{x \to -1} u(x)\)?

    • \(-\infty\)
    • \(\infty\)
    • The limit does not exist, because the two sides differ
    Answer

    \(-\infty\)

  2. What is \(\displaystyle\lim_{x \to 2^-} u(x)\)?

    • \(-\infty\)
    • \(\infty\)
    • It does not exist
    Answer

    \(-\infty\)

  3. Find \(\displaystyle\lim_{x \to 5^-} \frac{3 - x}{x - 5}\).

    • \(\infty\)
    • \(-\infty\)
    • 0
    • −2
    Answer

    \(\infty\)

Summary

You multiply the sign of the top near the zero of the bottom by the sign of the bottom on each side.

The product is the sign of the large height on that side, because the bottom is small and the top is not.

This gives each one-sided limit at a vertical asymptote and shows whether the two-sided limit exists.

This is one of four skills in topic 1.14 of Which & Why: AP Calculus.