Wemazi

Topic 1.7

Combine methods when necessary

Builds on using standard trigonometric limits. 7 minutes.

A quotient that needs two steps

The graph shows \(\dfrac{\tan 2x}{3x}\), with an orange dot at the input x. The open circle at 0 shows that the quotient has no value there.

The standard limits are those of \(\dfrac{\sin x}{x}\) and \(\dfrac{1-\cos x}{x}\), and a tangent fits neither. Rewrite \(\tan 2x\) as \(\dfrac{\sin 2x}{\cos 2x}\), and split the quotient into the factors \(\dfrac{\sin 2x}{3x}\) and \(\dfrac{1}{\cos 2x}\).

Press the button to draw the two factors. Their product is the chalk curve at every x except 0.

Move x toward 0. The first factor is \(\tfrac23\cdot\dfrac{\sin 2x}{2x}\), so it approaches \(\tfrac23\). The second approaches 1, so the quotient approaches \(\tfrac23\).

Rewrite, split, combine

The chalk curve is now any quotient near an input \(c\). Press Rewrite. The orange dashed curve covers the chalk curve, because a rewrite keeps every value near \(c\), so the limit does not change.

Press Split into pieces. The faint curves \(p\) and \(q\) are two pieces whose product is the quotient, and each orange dashed line marks a limit.

Move x toward \(c\). Each piece approaches its own line, so the quotient approaches the product of the two limits.

Notation

  1. Substitute \(c\) first. If the result is the red form \(\dfrac00\), no limit law gives an answer, so rewrite the expression before you use one.

  2. Choose a rewrite that produces pieces with known limits. Useful rewrites are an identity (an equation that is true for every input), a factorization, a common denominator or a conjugate. The orange dashed curve shows the rewritten expression.

    \[\begin{gathered}\tan u=\frac{\sin u}{\cos u}\\ 1-\cos 2u=2\sin^2 u\end{gathered}\]

  3. Split the expression into pieces. To give each sine its own angle in the denominator, multiply and divide by constants. The faint curves \(p\) and \(q\) show the two pieces.

  4. Apply the product law to the pieces. It holds only when every piece has a limit.

    \[\lim_{x\to c}\,\cq{p(x)}\,\cq{q(x)}=\cy{\lim p\cdot\lim q}\]

    A quotient law also needs the limit of the bottom to be different from 0.

  5. A rewrite needs to be true only for \(x\) near \(c\), because a limit does not use the value at \(c\).

When it fails

Take the limit of the top and the limit of the bottom separately. Both are 0, so the quotient law divides 0 by 0.

The quotient law needs a bottom whose limit is not 0, so it gives no answer here. Rewrite the expression first.

Example 1: an identity first

Find \(\displaystyle\lim_{x\to 0}\frac{1-\cos 2x}{x^2}\).

  1. Substitute \(x=0\). The top and the bottom are both 0, so the form is \(\dfrac00\). Neither standard limit fits, because the bottom is \(x^2\).

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

  2. The identity \(\cos 2x=1-2\sin^2 x\) rewrites the top with a sine. The orange dashed curve lies exactly over the original curve.

    \[\frac{1-\cos 2x}{x^2}=\frac{2\sin^2 x}{x^2}\]

  3. Split the quotient into factors, so that each sine has an \(x\) under it. The faint curve shows \(\dfrac{\sin x}{x}\).

    \[2\cdot\frac{\sin x}{x}\cdot\frac{\sin x}{x}\]

  4. Each factor \(\dfrac{\sin x}{x}\) approaches 1, so by the product law the limit is 2.

    \[2\cdot 1\cdot 1=\cy{2}\]

Example 2: split into pieces that have limits

Find \(\displaystyle\lim_{x\to 0} x\cot 3x\).

  1. Write the cotangent as cosine over sine. The faint curve shows \(\cot 3x\) alone, which rises without bound to the right of 0 and falls without bound to the left.

    \[x\cot 3x=x\cdot\frac{\cos 3x}{\sin 3x}\]

  2. A tempting wrong move: The first factor x approaches 0, so the product approaches 0.

    \[\lim_{x\to 0} x\cdot\lim_{x\to 0}\cot 3x=\cr{0}\]

    The product curve stays near height \(\tfrac13\), above the red line at 0. The factor \(\cot 3x\) has no limit at 0, so the product law does not apply.

  3. Rewrite the product as pieces that each have a limit. The piece \(\dfrac{x}{\sin 3x}\) is the reciprocal of \(\dfrac{\sin 3x}{x}\).

    \[\cos 3x\cdot\frac{x}{\sin 3x}\]

  4. The cosine approaches \(\cos 0=1\). The reciprocal piece approaches \(\tfrac13\), because \(\dfrac{\sin 3x}{x}\) approaches 3.

    \[1\cdot\frac13=\cy{\frac13}\]

Practice

Find \(\displaystyle\lim_{x\to 0}\frac{\tan 5x}{\sin 2x}\).

Move x toward 0 and read the height that the curve approaches. To find it exactly, write \(\tan 5x\) as sine over cosine and give each sine its own angle below it.

Solution

  1. Rewrite the tangent as a sine over a cosine.

    \[\frac{\sin 5x}{\cos 5x\,\sin 2x}\]

  2. Multiply and divide by \(5x\) and by \(2x\), so that each sine has its own angle in the denominator. The constants combine to \(\tfrac52\).

    \[\frac52\cdot\frac{\sin 5x}{5x}\cdot\frac{2x}{\sin 2x}\cdot\frac{1}{\cos 5x}\]

  3. The four pieces approach \(\tfrac52\), 1, 1 and 1.

    \[\frac52\cdot 1\cdot 1\cdot 1=\cy{\frac52}\]

Questions

The graph shows \(s(x)=\dfrac{\tan 4x}{x}\) and \(w(x)=\dfrac{1-\cos x}{x^2}\). Each curve has an open circle at \(x=0\), and the dashed vertical lines are asymptotes of \(s\).

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

    • 4
    • 1
    • 0
    • \(\dfrac14\)
    Answer

    4

  2. As \(x\) approaches 0, what does \(w(x)\) approach?

    • \(\dfrac12\)
    • 0
    • 1
    • 2
    Answer

    \(\dfrac12\)

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

    • 6
    • 5
    • 0
    • 1
    Answer

    6

Summary

You rewrite the expression, split it into pieces whose limits you know, and combine the limits with the limit laws.

The rewrite keeps the value at every x except 0, so the limit does not change.

Use this plan when substitution gives \(\dfrac00\) and no single standard limit fits.

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