Wemazi

Topic 1.8

Choose valid upper and lower bounds

Builds on using standard trigonometric limits. 6 minutes.

A wiggle between two parabolas

The curve is \(f(x)=x^2\cos\frac6x\). It oscillates faster and faster near 0, and the cosine part has no limit there, so the limit laws do not apply.

The cosine of any angle is between \(-1\) and 1. Multiplying by \(x^2\), which is never negative, keeps this order, so \(f(x)\) lies between \(-x^2\) and \(x^2\).

The faint dashed curves are the parabolas \(y=x^2\) and \(y=-x^2\). Drag the orange point to either side of 0. It stays on the orange segment between the parabolas, so the curve stays between them.

At x = 0.2, the lower parabola is at −0.04, the curve is at about 0.006, and the upper parabola is at 0.04.

Bounds that hold near c

The curve \(f\) is any function. The faint dashed curves \(b\) and \(a\) are a lower bound and an upper bound when \(f\) stays above \(b\) and below \(a\).

The dashed box marks a strip of inputs on both sides of \(c\). Inside the strip, \(f\) stays between \(b\) and \(a\), so \(b\) and \(a\) are valid bounds there.

Widen the strip. Far from c, the curve crosses the bounds, and the parts that cross turn red because the inequality is false there.

A limit uses only the inputs near \(c\), so the bounds need to hold only in a narrow strip, and not at \(c\) itself.

Notation

  1. To trap \(f\), you use two simpler functions. The lower bound \(b\) stays below \(f\), and the upper bound \(a\) stays above it.

    \[\cq{b(x)}\le f(x)\le \cq{a(x)}\]

  2. The inequalities must hold for every \(x\) near \(c\), on both sides of \(c\).

  3. A bound often comes from a known range. The cosine of any angle is between \(-1\) and 1.

    \[-1\le \cos u\le 1\]

  4. Multiplying the range by a factor that is never negative, such as \(x^2\) or \(|x|\), keeps the order of the inequality.

    \[\cq{-x^2}\le x^2\cos\tfrac6x\le \cq{x^2}\]

  5. If you add a constant, add it to all three parts. If you multiply by a negative number, reverse the order of the inequality.

When it fails

Make the upper parabola smaller. Where \(f\) rises above it, the curve turns red, because the inequality is false there.

An upper bound is valid only if \(f\) stays below it at every input close to 0. Here \(f\) rises above the smaller parabola at inputs as close to 0 as you like, so the smaller parabola is not a valid bound.

Example 1: a shifted wiggle

Find a lower bound and an upper bound for \(f(x)=(x-2)^2\cos\dfrac{5}{x-2}\), valid for all \(x\) near 2 with \(x\ne 2\).

  1. The cosine of any angle is between \(-1\) and 1. Here the angle is \(\frac{5}{x-2}\), which has a value at every \(x\ne2\).

    \[-1\le \cos\frac{5}{x-2}\le 1\]

  2. The factor \((x-2)^2\) is never negative, so multiplying every part by it keeps the order.

    \[-(x-2)^2\le f(x)\le (x-2)^2\]

  3. The factor is positive on both sides of 2, so the inequalities are true on both sides.

Example 2: a factor that can be negative

Find a lower bound and an upper bound for \(f(x)=x\sin\dfrac4x\), valid for all \(x\) near 0 with \(x\ne 0\).

  1. The sine of any angle is between \(-1\) and 1.

    \[-1\le \sin\frac4x\le 1\]

  2. A tempting wrong move: Multiply by \(x\).

    \[\cr{-x\le f(x)\le x}\]

    On the left of 0, the line \(y=-x\) is above \(y=x\), so no curve fits between them and the left half is red. At \(x=-0.1\), the claim says that \(f\) is at least 0.1, but \(f\) is about 0.07.

  3. Multiplying by \(x\) reverses the order where \(x\) is negative, so use absolute values instead. The absolute value of the sine is at most 1, so \(|f(x)|\) is at most \(|x|\), which means that \(f\) lies between \(-|x|\) and \(|x|\).

    \[\begin{gathered}|f(x)|=|x|\left|\sin\tfrac4x\right|\le |x|\\ -|x|\le x\sin\tfrac4x\le |x|\end{gathered}\]

Practice

Find a lower bound and an upper bound for \(f(x)=1-x^2\sin\dfrac4x\), valid for all \(x\) near 0 with \(x\ne 0\).

The curve is \(x^2\sin\frac4x\) reflected in the x-axis and moved up by 1. Bound \(x^2\sin\frac4x\) first, then do the same to the bounds.

Solution

  1. The sine is between \(-1\) and 1, and \(x^2\) is never negative, so \(x^2\sin\frac4x\) lies between \(-x^2\) and \(x^2\).

    \[-x^2\le x^2\sin\frac4x\le x^2\]

  2. Multiplying by \(-1\) reverses the order, so the bounds \(-x^2\) and \(x^2\) swap places.

    \[-x^2\le -x^2\sin\frac4x\le x^2\]

  3. Add 1 to all three parts.

    \[\cq{1-x^2}\le f(x)\le \cq{1+x^2}\]

Questions

The graph shows \(g(x)=x^2\sin\dfrac5x\) near 0, with five dashed candidates for a bound.

  1. Which dashed curve is a valid upper bound for \(g\) near 0?

    • \(y=x^2\)
    • \(y=-x^2\)
    • \(y=\tfrac12x^2\)
    • \(y=0\)
    Answer

    \(y=x^2\)

  2. For which \(x\) near 0 is the line \(y=x\) above the graph of \(g\)?

    • Only for \(x>0\)
    • Only for \(x<0\)
    • For every \(x\) near 0
    • For no \(x\) near 0
    Answer

    Only for \(x>0\)

  3. Which bounds are valid for \(f(x)=x^2\left(4-\cos\dfrac1x\right)\) for all \(x\) near 0 with \(x\ne 0\)?

    • \(3x^2\le f(x)\le 5x^2\)
    • \(5x^2\le f(x)\le 3x^2\)
    • \(-x^2\le f(x)\le x^2\)
    • \(4x^2\le f(x)\le 5x^2\)
    Answer

    \(3x^2\le f(x)\le 5x^2\)

Summary

You write a lower bound and an upper bound from a known range, such as the cosine between −1 and 1.

The function must stay between the bounds on both sides of c, so you multiply only by factors that are never negative, such as x².

Use bounds when a function oscillates and the limit laws do not apply.

This is one of four skills in topic 1.8 of Which & Why: AP Calculus.