Rewrite other indeterminate forms
A product of a shrinking and a growing factor
As \(x\) approaches 0 from the right, \(x^2\) approaches 0 and \(\ln x\) falls without bound, so \(x^2\ln x\) has the form \(0\cdot\infty\). The upper panel shows \(x^2\) as a solid curve and \(\ln x\) as a dashed curve, and the lower panel shows their product.
Drag \(x\) toward 0. At \(x=0.1\) the factors are 0.01 and about \(-2.30\), but their product is only about \(-0.023\).
L'Hôpital's rule needs a quotient, so press Write as a quotient. It moves \(x^2\) to the bottom as its reciprocal \(\frac{1}{x^2}\), and the product becomes \(\dfrac{\ln x}{1/x^2}\). The curve in the lower panel does not move, but the form changes to \(\frac{\infty}{\infty}\).
A rewrite changes the form, not the value
The upper panel shows a factor that approaches 0 and a factor that falls without bound, and the lower panel shows their product. Press the first button to write the product as a quotient.
The curve in the lower panel does not move, because both expressions have the same value. Only the form changes, from \(0\cdot\infty\) to \(\frac{\infty}{\infty}\).
A difference of two terms that both grow without bound has the form \(\infty-\infty\). Press the second button to write it as one fraction over a common denominator, which has the form \(\frac{0}{0}\).
A power \(y=f^g\) has a variable exponent, and the logarithm turns it into the product \(\ln y=g\ln f\), which you write as a quotient.
Press the third button. The solid curve is \(\ln y\), which approaches the orange line at \(L\), and the dashed curve is \(y\), which approaches \(e^L\).
Notation
For a product, move one factor to the bottom as its reciprocal. Choose the factor that gives the simpler quotient after you differentiate.
\[f\cdot g=\dfrac{g}{1/f}=\dfrac{f}{1/g}\]For a difference of fractions, combine the two terms into one fraction.
\[\dfrac{1}{u}-\dfrac{1}{v}=\dfrac{v-u}{uv}\]For a power with the form \(1^\infty\), \(0^0\) or \(\infty^0\), let \(y=f^g\) and take the logarithm. If \(\ln y\) approaches \(L\), then \(y\) approaches \(e^{L}\), because \(e^t\) is continuous.
\[\begin{aligned} \ln y &= g\ln f \\ \ln y\to L\ &\Rightarrow\ y\to e^{L} \end{aligned}\]A base that approaches 0 with an exponent that grows without bound is not indeterminate. The power approaches 0.
When it fails
Write \(x^2\ln x\) the other way, as \(\dfrac{x^2}{1/\ln x}\). The derivatives are \(2x\) and \(-\dfrac{1}{x\ln^2 x}\).
The new quotient is \(-2x^2\ln^2 x\), the dashed curve in the lower panel. It has the form \(0\cdot\infty\) again and is harder than the product, so put \(\ln x\) on top instead, because its derivative \(\frac1x\) is simple.
Example 1: A power with a base near 1
Find \(\displaystyle\lim_{x\to \infty}\left(1+\frac{2}{x}\right)^x\).
The base approaches 1 and the exponent grows without bound, so the form is \(1^\infty\). Let \(y\) be the power and take the logarithm. The solid curve in the lower panel shows \(\ln y\).
\[\ln y=x\ln\left(1+\frac{2}{x}\right)\]This product has the form \(\infty\cdot0\). Move \(x\) to the bottom as \(\frac{1}{x}\), and the form is now \(\frac{0}{0}\).
\[\ln y=\dfrac{\ln(1+2/x)}{1/x}\]Differentiate the top and the bottom, each by itself. The factor \(-\frac{1}{x^2}\) cancels, and the quotient approaches 2.
\[\dfrac{\dfrac{-2/x^2}{1+2/x}}{-1/x^2}=\dfrac{2}{1+2/x}\to 2\]Since \(\ln y\) approaches 2, \(y\) approaches \(e^2\). The dashed curve shows \(y\). As a check, \(e^2\approx7.39\).
\[\lim_{x\to\infty}\left(1+\frac{2}{x}\right)^x=e^{2}\]
Example 2: A difference of two fractions
Find \(\displaystyle\lim_{x\to 0^+}\left(\frac{1}{x}-\frac{1}{e^x-1}\right)\).
As \(x\) approaches 0 from the right, \(\frac{1}{x}\) and \(\frac{1}{e^x-1}\) both grow without bound. The form is \(\infty-\infty\).
A tempting wrong move: The two terms are equally large, so they cancel. The limit is 0.
\[\infty-\infty=0\]Two terms that grow without bound need not cancel, because their difference depends on how fast each grows. At \(x=0.1\) the difference is about 0.492, and the curve stays well above the red line at 0.
Combine the two terms over a common denominator. The form is now \(\frac{0}{0}\).
\[\frac{1}{x}-\frac{1}{e^x-1}=\frac{e^x-1-x}{x(e^x-1)}\]Differentiate the top and the bottom. Both are still 0 at \(x=0\), so differentiate again.
\[\dfrac{e^x-1}{e^x-1+xe^x}\ \longrightarrow\ \dfrac{e^x}{2e^x+xe^x}\]Substitute \(x=0\). The top is 1 and the bottom is 2, so the limit is \(\frac{1}{2}\).
\[\lim_{x\to0^+}\left(\frac1x-\frac{1}{e^x-1}\right)=\frac{1}{2}\]
Practice
Find \(\displaystyle\lim_{x\to \infty}x\left(e^{1/x}-1\right)\).
In the upper panel, the solid factor x grows without bound. Move it to the bottom as its reciprocal.
Solution
The factor \(x\) grows without bound and the factor \(e^{1/x}-1\) approaches 0, so the form is \(\infty\cdot0\).
Move \(x\) to the bottom as \(\frac{1}{x}\). The form is now \(\frac{0}{0}\).
\[x\left(e^{1/x}-1\right)=\dfrac{e^{1/x}-1}{1/x}\]Differentiate the top and the bottom. The factor \(-\frac{1}{x^2}\) cancels. As \(x\) grows, \(e^{1/x}\) approaches 1, so the limit is 1.
\[\dfrac{e^{1/x}\cdot(-1/x^2)}{-1/x^2}=e^{1/x}\to 1\]
Questions
Questions 1 and 2 use the curve \(y=xe^{2x}\), which is the product of the factors \(x\) and \(e^{2x}\).
As \(x\) moves far to the left, the factor \(x\) falls without bound and the factor \(e^{2x}\) approaches 0. What does \(y\) approach?
- 0
- \(-\infty\)
- About −0.18
- The limit does not exist.
Answer
0
As \(x\) moves far to the right, both factors grow without bound. What does \(y\) do?
- It grows without bound.
- It approaches 0, as on the left.
- It is indeterminate, so it needs L'Hôpital's rule.
- It approaches −0.18.
Answer
It grows without bound.
What is \(\displaystyle\lim_{x\to0^+}x^{2x}\)?
- 1
- 0
- \(e^2\)
- It does not exist.
Answer
1
Summary
You rewrite a product, a difference or a power as a quotient of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), and then you apply L'Hôpital's rule.
This works because a rewrite does not change the value, and the logarithm turns a power into a product.
You use a rewrite when substitution gives \(0\cdot\infty\), \(\infty-\infty\), \(1^\infty\), \(0^0\) or \(\infty^0\).
This is one of four skills in topic 4.7 of Which & Why: AP Calculus.