Wemazi

Topic 4.7

Differentiate numerator and denominator separately

Builds on checking the form and conditions for L'Hôpital's rule. 7 minutes.

Two heights that shrink at different rates

Substitute \(x=0\) into \(\dfrac{e^x-\cos x}{\sin 2x}\). The top is 0 and the bottom is 0, so the form is \(\frac{0}{0}\).

The upper panel shows the top as a solid curve and the bottom as a dashed curve. The orange bars show their heights at the marked \(x\). Drag \(x\) toward 0.

Both bars shrink toward 0, but the bottom bar shrinks about twice as fast. The lower panel shows the quotient of the two heights, which moves toward \(\frac12\).

The rate at which a bar shrinks is the derivative of its function at 0. Show the tangent lines. The top has rate \(f'(0)=1\) and the bottom has rate \(g'(0)=2\), so the quotient of the rates is also \(\frac12\).

The quotient of the rates

The upper panel shows two functions \(f\) and \(g\) that are both 0 at \(a\), with orange tangent lines at \(a\). The lower panel shows their quotient.

Drag the slider to tilt \(g\). The quotient curve changes, but as \(x\) nears \(a\) it approaches the orange line at the ratio of the two slopes.

Each slope is a derivative, so you differentiate \(f\) alone and \(g\) alone, and the limit is \(\frac{\cy{f'(a)}}{\cy{g'(a)}}\).

Show the slope of the quotient curve. The red tangent line has the slope of \(\frac{f}{g}\), a different number. The limit is the height of the quotient curve, not its slope.

Notation

  1. Differentiate the top by itself. Differentiate the bottom by itself. Then take the limit of the first derivative over the second.

    \[\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{\cy{f'(x)}}{\cy{g'(x)}}\]

    This holds if the form is \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) and the new limit exists or is infinite.

  2. This is not the quotient rule. The quotient rule gives the derivative of the single function \(\frac{f}{g}\), a different expression.

    \[\left(\frac{f}{g}\right)'=\cr{\frac{f'g-fg'}{g^2}}\ne\cy{\frac{f'}{g'}}\]

  3. Each derivative is a separate calculation, so use the product rule, the chain rule or any other rule inside it.

When it fails

The top \(\tan x\) and the bottom \(x^2\) both approach 0. Differentiating each one separately gives \(\dfrac{\sec^2 x}{2x}\), the dashed curve in the lower panel.

From the right of 0 this new quotient grows without bound, and from the left it falls without bound. The new limit does not exist, so the rule gives no answer. The solid curve, the original quotient, splits in the same way.

Example 1: A fourth power over a logarithm

Find \(\displaystyle\lim_{x\to 1}\frac{x^4-1}{\ln x}\).

  1. At \(x=1\) the top is 0 and the bottom is 0, so the form is \(\frac{0}{0}\) and the rule applies.

    \[1^4-1=0,\qquad \ln 1=0\]

  2. The top \(x^4-1\) has derivative \(4x^3\), and the bottom \(\ln x\) has derivative \(\frac1x\). At \(x=1\) these are 4 and 1, the slopes of the orange tangent lines.

    \[\frac{(x^4-1)'}{(\ln x)'}=\frac{\cy{4x^3}}{\cy{1/x}}\]

  3. Substitute \(x=1\) into the new quotient, which gives \(\frac41=4\). The quotient curve approaches the orange line at height 4.

    \[\lim_{x\to1}\frac{4x^3}{1/x}=\lim_{x\to1}4x^4=4\]

Example 2: A logarithm over x

Find \(\displaystyle\lim_{x\to 0}\frac{\ln(1+3x)}{x}\).

  1. At \(x=0\) the top is \(\ln 1=0\) and the bottom is 0. The form is \(\frac{0}{0}\).

  2. A tempting wrong move: Treat the quotient as one function. Use the quotient rule and find the limit of the result.

    \[\lim_{x\to0}\left(\frac{\ln(1+3x)}{x}\right)'=-\frac92\]

    The red tangent line shows that \(-\frac92\) is the slope of the quotient curve at \(x=0\), not its height. At \(x=0.1\) the curve is at height about 2.62, so \(-\frac92\) cannot be the limit.

  3. Differentiate the top and the bottom separately. By the chain rule, \(\ln(1+3x)\) has derivative \(\frac{3}{1+3x}\), and \(x\) has derivative 1. The limit is 3, the height of the orange line.

    \[\lim_{x\to0}\frac{3/(1+3x)}{1}=3\]

Practice

Find \(\displaystyle\lim_{x\to 0}\frac{\sqrt{1+8x}-1}{\sin 5x}\).

Differentiate the top by itself and the bottom by itself. Each one needs the chain rule. Then compare the two slopes at the origin.

Solution

  1. At \(x=0\) the top is \(\sqrt1-1=0\) and the bottom is \(\sin 0=0\). The form is \(\frac{0}{0}\).

    \[\sqrt{1+8\cdot0}-1=0,\qquad\sin0=0\]

  2. Differentiate the top and the bottom separately. At the origin the derivatives are 4 and 5, the slopes of the orange tangent lines.

    \[\begin{aligned} \frac{d}{dx}\sqrt{1+8x} &= \frac{4}{\sqrt{1+8x}} \\ \frac{d}{dx}\sin 5x &= 5\cos 5x \end{aligned}\]

  3. Substitute \(x=0\) into the new quotient, which gives \(\frac45\).

    \[\lim_{x\to0}\frac{4/\sqrt{1+8x}}{5\cos5x}=\frac45\]

Questions

Questions 1 and 2 use \(f(x)=6x-x^3\), the solid curve, and \(g(x)=e^{2x}-1\), the dashed curve.

  1. The orange tangent lines at the origin have slopes 6 and 2. What is \(\displaystyle\lim_{x\to0}\frac{f(x)}{g(x)}\)?

    • 3
    • \(\frac13\)
    • 4
    • 12
    Answer

    3

  2. At \(x=0.2\), the marked heights are \(f(0.2)\approx1.19\) and \(g(0.2)\approx0.49\). Their quotient is about 2.4. Which value does \(\frac{f(x)}{g(x)}\) approach as \(x\to0\)?

    • 3
    • 2.4
    • 0
    • 1
    Answer

    3

  3. What is \(\displaystyle\lim_{x\to0}\frac{e^{5x}-1}{\ln(1+2x)}\)?

    • \(\frac52\)
    • \(\frac25\)
    • \(\frac{35}{4}\)
    • 0
    Answer

    \(\frac52\)

Summary

You differentiate the top and the bottom separately, then take the limit of the first derivative over the second.

This works because each function shrinks at the rate of its derivative, so the quotient of the heights approaches the quotient of the rates.

You use it after the form is \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), and you do not use the quotient rule on the original fraction.

This is one of four skills in topic 4.7 of Which & Why: AP Calculus.