Wemazi

Topic 1.15

Handle positive and negative infinity separately

Builds on finding the end behavior of a rational function. 7 minutes.

Two ends, two answers

The curve shows \(g(x) = \dfrac{6x}{\sqrt{9x^2 + 1}}\).

Move the orange point far to the right. The height approaches 2, and a dashed orange line marks it.

Move the point far to the left. The height approaches −2, and a second dashed orange line marks it.

The bottom is a square root, so it is positive for every x. The top has the sign of x, so g is positive on the right and negative on the left.

Each end has its own height

The curve is any function with a left end and a right end. Each end approaches its own dashed line, at the height R on the right and the height L on the left.

Move either slider. Only the matching end of the curve changes, because one end does not decide the other. The two heights can be equal, opposite, or different.

A limit at infinity describes one end only, so you find it once for each end.

Notation

  1. Write the two ends as two limits. The symbol \(x \to -\infty\) means that x is negative and its size grows without bound.

    \[\lim_{x \to \infty} g(x) \quad\text{and}\quad \lim_{x \to -\infty} g(x)\]

  2. A square root is never negative, so \(\sqrt{x^2}\) is the absolute value of x, not x itself.

    \[\sqrt{x^2} = |x| = \begin{cases} x & x \ge 0 \\ -x & x < 0 \end{cases}\]

  3. To compare the bottom with x, factor \(x^2\) out of the root, which turns \(\sqrt{x^2}\) into \(|x|\).

    \[\sqrt{9x^2 + 1} = |x| \sqrt{9 + \cq{1/x^2}}\]

  4. At the right end x is positive, so \(|x|\) equals x. The factor x cancels from the top and the bottom, and \(1/x^2\) approaches 0.

    \[\frac{6x}{x\sqrt{9 + 1/x^2}} \to \frac{6}{\sqrt 9} = \cy{2}\]

  5. At the left end x is negative, so \(|x|\) equals \(-x\). The bottom has the opposite sign from the top, so the limit is negative.

    \[\frac{6x}{-x\sqrt{9 + 1/x^2}} \to \frac{6}{-\sqrt 9} = \cy{-2}\]

    An absolute value without a root follows the same rule: replace \(|x|\) by \(-x\) when x is negative.

When it fails

Suppose you replace \(|x|\) by x on the left end. The calculation gives 2 again, but 6x is negative and the root is positive, so the curve on the left lies below the x-axis.

A curve below the axis cannot approach a positive height, so the red dashed line at 2 stays far from the curve. The limit on the left is −2.

Example 1: an absolute value

Find the limit of \(f(x) = \dfrac{5x}{|x| + 2}\) as \(x \to \infty\) and as \(x \to -\infty\).

  1. At the right end x is positive, so \(|x|\) equals x. Divide the top and the bottom by x, and \(2/x\) approaches 0.

    \[\frac{5x}{x + 2} = \frac{5}{1 + \cq{2/x}} \to \cy{5}\]

  2. At the left end x is negative, so \(|x|\) equals \(-x\) and the bottom is \(2 - x\).

    \[\frac{5x}{2 - x}\]

  3. Divide the top and the bottom by x, which turns the bottom into \(2/x - 1\).

    \[\frac{5}{\cq{2/x} - 1} \to \frac{5}{-1} = \cy{-5}\]

Example 2: a root with a shifted top

Find both end limits of \(h(x) = \dfrac{3 - x}{\sqrt{x^2 + 8}}\).

  1. Factor \(x^2\) out of the root. The root of \(x^2\) is \(|x|\).

    \[\sqrt{x^2 + 8} = |x| \sqrt{1 + \cq{8/x^2}}\]

  2. At the right end \(|x|\) equals x, so the bottom is \(x\sqrt{1 + 8/x^2}\). Divide the top and the bottom by x.

    \[\frac{3/x - 1}{\sqrt{1 + 8/x^2}} \to \cy{-1}\]

  3. A tempting wrong move: At the left end, use the same steps. The limit is −1 again.

    \[\lim_{x \to -\infty} h(x) = \cr{-1}\]

    For negative x, the top \(3 - x\) and the root are both positive, so the left end of the curve lies above the axis. The red line at −1 is below the axis, far from the curve.

  4. At the left end \(|x|\) equals \(-x\), so the bottom is \(-x\sqrt{1 + 8/x^2}\). Divide the top and the bottom by \(-x\).

    \[\frac{-3/x + 1}{\sqrt{1 + 8/x^2}} \to \cy{1}\]

Practice

Find both end limits of \(p(x) = 7 + 6x - x^3\).

The term with the highest power decides each end. Check the sign of that term on each side of the picture.

Solution

  1. The term \(-x^3\) has the highest power. Factor it out.

    \[p(x) = -x^3 \left(1 - \cq{6/x^2} - \cq{7/x^3}\right)\]

  2. For large positive x, \(x^3\) is large and positive, so \(-x^3\) is large and negative. The bracket approaches 1, so p falls without bound.

    \[\lim_{x \to \infty} p(x) = \cy{-\infty}\]

  3. For large negative x, \(x^3\) is large and negative, so \(-x^3\) is large and positive. The bracket still approaches 1, so p rises without bound.

    \[\lim_{x \to -\infty} p(x) = \cy{\infty}\]

Questions

The graph shows \(q(x) = 3 + \dfrac{2x}{|x| + 1}\) for x from \(-40\) to 40.

  1. As x grows to the right, which height does the curve approach?

    • 5
    • 1
    • 3
    • 2
    Answer

    5

  2. As x grows to the left, which height does the curve approach?

    • 1
    • 5
    • 3
    • −2
    Answer

    1

  3. What is \(\displaystyle\lim_{x \to -\infty} \frac{5x - 1}{\sqrt{4x^2 + 3}}\)?

    • −5/2
    • 5/2
    • −5/4
    • The limit does not exist
    Answer

    −5/2

Summary

You find the limit at infinity and the limit at negative infinity separately, because each end of a curve has its own height.

The two limits can differ because \(\sqrt{x^2}\) and \(|x|\) both equal \(-x\) on the left end, where x is negative.

Treat the ends separately whenever a formula has a root, an absolute value, or an odd power of x.

This is one of four skills in topic 1.15 of Which & Why: AP Calculus.