Wemazi

Topic 1.7

Use standard trigonometric limits

Builds on choosing a useful first step. 7 minutes.

An arc and a segment on a circle

A point moves on a circle of radius 1. The angle x is in radians, which means that the orange arc from the start of the circle to the point has length x.

The vertical segment from the point down to the horizontal axis has length \(\sin x\). Drag x toward 0. The segment and the arc become almost equal: at \(x = 0.1\), the arc is 0.1 and the segment is about 0.0998.

The graph on the right plots the ratio of the segment to the arc, \(\dfrac{\sin x}{x}\). The ratio approaches 1 from both sides of 0, and the open circle at 0 shows that the ratio has no value there.

The ratio near zero

Drag x to the left of 0. The arc and the segment both point downward, so both lengths count as negative. The ratio of two negative numbers is positive, so the graph has the same height at x and at −x.

Shrink x toward 0. Near the start of the arc, the circle is almost a vertical line, so the arc and the segment have nearly the same length. The ratio therefore approaches 1 from either side of 0.

This holds only when the angle is in radians, because only then is the arc length equal to the angle.

Notation

  1. If the angle \(\theta\) is in radians and approaches 0, the ratio of its sine to the angle approaches 1. The figure uses \(x\) for the angle.

    \[\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=\cy{1}\]

  2. The second standard limit is for cosine. Multiply the top and the bottom by \(1+\cos\theta\), so the top becomes \(1-\cos^2\theta=\sin^2\theta\). The second factor then approaches \(\tfrac{0}{2}\), and the orange curve \(\dfrac{1-\cos x}{x}\) approaches height 0.

    \[\begin{gathered}\frac{1-\cos\theta}{\theta}=\frac{\sin\theta}{\theta}\cdot\frac{\sin\theta}{1+\cos\theta}\\ \to \cy{1}\cdot\cy{0}=\cy{0}\end{gathered}\]

  3. The angle inside the sine must match the denominator. If they differ by a constant factor \(k\), multiply and divide by \(k\). The figure uses \(k=3\), and the faint curve shows the matching fraction \(\dfrac{\sin 3x}{3x}\).

    \[\frac{\sin kx}{x}=k\cdot\frac{\sin kx}{kx}\]

  4. As \(x\) approaches 0, \(kx\) also approaches 0, so the fraction \(\dfrac{\sin kx}{kx}\) approaches 1 and the limit is \(k\).

    \[\lim_{x\to 0}\frac{\sin kx}{x}=\cy{k}\]

  5. The expression \(\sin kx\) is not \(k\sin x\), because the constant \(k\) multiplies the angle and not the sine. The red curve shows \(\dfrac{k\sin x}{x}\), which differs from the chalk curve away from 0.

When it fails

In degrees, the angle number x is about 57.3 times the arc length, so the arc is no longer x. The segment is still the sine of the angle.

The ratio \(\dfrac{\sin x}{x}\) then approaches \(\tfrac{\pi}{180}\), about 0.0175, and not 1, so the limits above need radians.

Example 1: a scale factor

Find \(\displaystyle\lim_{x\to 0}\frac{\sin 5x}{x}\).

  1. The sine contains \(5x\), but the denominator is \(x\), and the standard limit needs the same expression in both places.

    \[\frac{\sin 5x}{x}\]

  2. Multiply and divide by 5, which does not change the value because \(\tfrac55=1\). The denominator is now \(5x\), and the faint curve shows the fraction \(\dfrac{\sin 5x}{5x}\).

    \[5\cdot\frac{\sin 5x}{5x}\]

  3. As \(x\) approaches 0, the angle \(5x\) also approaches 0. The fraction approaches 1, so the limit is 5 times 1.

    \[5\cdot 1=\cy{5}\]

Example 2: the angle is not x

Find \(\displaystyle\lim_{x\to 0}\frac{\sin(x^2)}{x}\).

  1. The angle in the sine is \(x^2\), but the denominator is \(x\), so the two do not match.

    \[\frac{\sin(x^2)}{x}\]

  2. A tempting wrong move: This has the form sine over x, so the limit is 1.

    \[\lim_{x\to 0}\frac{\sin(x^2)}{x}=\cr{1}\]

    The curve near 0 is far below the red line at height 1. At x = 0.1, the quotient is about 0.1, and it becomes smaller as x shrinks.

  3. The angle is \(x^2\), so the denominator must be \(x^2\). Multiply and divide by \(x\). The faint curve shows \(\dfrac{\sin(x^2)}{x^2}\).

    \[x\cdot\frac{\sin(x^2)}{x^2}\]

  4. As \(x\) approaches 0, the angle \(x^2\) approaches 0, so the fraction approaches 1. The factor \(x\) approaches 0, so the product approaches 0.

    \[0\cdot 1=\cy{0}\]

Practice

Find \(\displaystyle\lim_{x\to 0}\frac{x}{\sin 4x}\).

Find the limit of the reciprocal \(\dfrac{\sin 4x}{x}\). Then drag x toward 0 and compare the height of the curve.

Solution

  1. Write the reciprocal \(\dfrac{\sin 4x}{x}\), and multiply and divide by 4. The fraction \(\dfrac{\sin 4x}{4x}\) approaches 1, so the reciprocal approaches 4.

    \[\frac{\sin 4x}{x}=4\cdot\frac{\sin 4x}{4x}\to 4\]

  2. Because 4 is not 0, the original quotient approaches the reciprocal \(\tfrac14\), and the dot on the curve approaches height \(\tfrac14\).

    \[\frac{x}{\sin 4x}\to\cy{\frac14}\]

Questions

The graph shows \(h(x)=\dfrac{\sin 3x}{x}\) and \(d(x)=\dfrac{\sin x}{x}\) when \(x\) is measured in degrees. Each curve has an open circle at \(x=0\). The faint dashed line is at height 1.

  1. As \(x\) approaches 0, what does \(h(x)\) approach?

    • 3
    • 1
    • 0
    • \(\dfrac13\)
    Answer

    3

  2. What does the graph show about the limit of \(d(x)\) as \(x\) approaches 0?

    • The limit is about 0.0175, not 1, because x is in degrees
    • The limit is 1, as for every angle
    • The limit is about 57.3
    • The limit does not exist
    Answer

    The limit is about 0.0175, not 1, because x is in degrees

  3. Find \(\displaystyle\lim_{x\to 0}\frac{\sin 8x}{2x}\).

    • 4
    • 1
    • 8
    • \(\dfrac14\)
    Answer

    4

Summary

You multiply and divide by the same factor so that the angle inside the sine matches the denominator.

The fraction \(\dfrac{\sin\theta}{\theta}\) then approaches 1, because in radians the arc and the segment have nearly the same length when the angle is small.

These limits hold only when the angle is in radians and approaches 0.

This is one of four skills in topic 1.7 of Which & Why: AP Calculus.