Check the form again after each step
A quotient that is still 0 over 0 after one step
Substitute \(x=0\) into \(\dfrac{e^x-1-x-\frac{x^2}{2}}{x^3}\). The top is 0 and the bottom is 0, so the form is \(\frac{0}{0}\).
The figure shows the top as a solid curve and the bottom as a dashed curve, and the orange dots mark their heights at x = 0. Press the button to differentiate each one by itself.
After the first step, the top is \(e^x-1-x\) and the bottom is \(3x^2\). Both dots are still at height 0, so the form is still \(\frac{0}{0}\), and it is the same after the second step.
After the third step, the top is \(e^x\) and the bottom is 6, so the dots are at heights 1 and 6. Neither is 0, so the form is no longer indeterminate, and the limit is \(\frac{e^0}{6}=\frac{1}{6}\).
The curve below is the original quotient, and it approaches the orange line at height \(\frac{1}{6}\).
Test the form after every differentiation
The figure shows the top and the bottom after each differentiation, and the orange dots show their heights at a.
While both dots are at 0, the form is still \(\frac{0}{0}\), so you differentiate again, and the new quotient keeps the limit of the original.
If one dot leaves 0, the form is not indeterminate, so you substitute a into the new quotient to read the limit.
Notation
After each differentiation, find the form of the new quotient. While it is \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), the rule gives the next equal limit.
\[\begin{aligned} \lim_{x\to a}\frac{f(x)}{g(x)} &= \lim_{x\to a}\frac{f'(x)}{g'(x)} \\ &= \lim_{x\to a}\frac{f''(x)}{g''(x)} = \cdots \end{aligned}\]Each equal sign needs the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) at the step before it.
Stop at the first quotient that is not indeterminate. Substitute a to read the limit, which may be a number or may grow without bound.
Simplify the new quotient before you differentiate again. A common factor may cancel and end the indeterminate form.
When it fails
Both \(x\) and \(\sqrt{x^2+1}\) grow without bound, so the form of \(\dfrac{x}{\sqrt{x^2+1}}\) is \(\frac{\infty}{\infty}\). One step gives the quotient \(\dfrac{\sqrt{x^2+1}}{x}\), and a second step returns to \(\dfrac{x}{\sqrt{x^2+1}}\). The form is \(\frac{\infty}{\infty}\) each time, so the rule makes no progress.
Divide the top and the bottom by \(x\) instead. For \(x>0\) this gives \(\dfrac{1}{\sqrt{1+1/x^2}}\), which approaches 1, the height of the orange line.
Example 1: A cosine over x squared
Find \(\displaystyle\lim_{x\to 0}\frac{1-\cos 3x}{x^2}\).
At \(x=0\) the top and the bottom are both 0, so both orange dots are at height 0 and the form is \(\frac{0}{0}\).
\[1-\cos 0=0,\qquad 0^2=0\]Differentiate each one by itself.
\[\frac{(1-\cos3x)'}{(x^2)'}=\frac{3\sin3x}{2x}\]At \(x=0\) the new top \(3\sin 0\) and the new bottom \(2\cdot 0\) are both 0, so the form is still \(\frac{0}{0}\). Differentiate again.
\[\frac{(3\sin3x)'}{(2x)'}=\frac{9\cos3x}{2}\]At \(x=0\) the top is \(9\cos 0=9\) and the bottom is 2. The form is no longer indeterminate, so the limit is \(\frac92\), the height of the orange line below.
\[\lim_{x\to0}\frac{1-\cos3x}{x^2}=\frac{9\cos 0}{2}=\frac92\]
Example 2: A quotient that stops being 0 over 0
Find \(\displaystyle\lim_{x\to 0}\frac{e^{2x}-1-2x}{x+x^2}\).
At \(x=0\) the top is \(e^0-1-0=0\) and the bottom is 0. The form is \(\frac{0}{0}\).
Differentiate each one by itself.
\[\frac{(e^{2x}-1-2x)'}{(x+x^2)'}=\frac{2e^{2x}-2}{1+2x}\]A tempting wrong move: The new top is still 0 at \(x=0\), so differentiate again.
\[\lim_{x\to0}\frac{4e^{2x}}{2}=2\]The new bottom is \(1+2\cdot0=1\), not 0, so the form is not \(\frac{0}{0}\). The original quotient below approaches height 0 at \(x=0\), far below the red line at height 2.
The new bottom approaches 1, so the quotient is not indeterminate. Substitute \(x=0\) into it.
\[\lim_{x\to0}\frac{2e^{2x}-2}{1+2x}=\frac{0}{1}=0\]
Practice
Find \(\displaystyle\lim_{x\to \infty}\frac{x^2}{e^{3x}}\).
Move the slider and read the tag in the upper panel. Stop at the first quotient where the top or the bottom does not grow without bound.
Solution
Both \(x^2\) and \(e^{3x}\) grow without bound, so the form is \(\frac{\infty}{\infty}\). Differentiate each one by itself.
\[\frac{2x}{3e^{3x}}\]The new top and the new bottom still grow without bound, so the form is still \(\frac{\infty}{\infty}\). Differentiate again.
\[\frac{2}{9e^{3x}}\]The top stays 2 while the bottom grows without bound, so the form is not indeterminate and the quotient approaches 0.
\[\lim_{x\to\infty}\frac{x^2}{e^{3x}}=\lim_{x\to\infty}\frac{2}{9e^{3x}}=0\]
Questions
Questions 1 and 2 follow \(\displaystyle\lim_{x\to0}\frac{\sin x-x}{x^2}\) through two differentiations. The solid curve is the top, and the dashed curve is the bottom.
After the first differentiation, the top is \(\cos x-1\) and the bottom is \(2x\). What is the next move?
- Differentiate again, because both are 0 at \(x=0\).
- Stop, because the top is 0, so the limit is 0.
- Stop, because the form is \(\frac{\infty}{\infty}\).
- Stop, because the limit does not exist.
Answer
Differentiate again, because both are 0 at \(x=0\).
After the second differentiation, the top is \(-\sin x\) and the bottom is 2. What is the limit?
- 0
- −1
- 2
- The limit does not exist.
Answer
0
What is \(\displaystyle\lim_{x\to\pi/2}\frac{1-\sin x}{\cos^2 x}\)?
- \(\frac12\)
- 0
- 1
- It does not exist.
Answer
\(\frac12\)
Summary
After each differentiation, you find the form of the new quotient, and you differentiate again only while it is \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
This works because each equal sign in the rule needs that form at the step before it.
You stop at the first quotient that is not indeterminate and substitute to read the limit.
This is one of four skills in topic 4.7 of Which & Why: AP Calculus.