Zero Times Infinity Converter

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About This MicroSim

L'Hospital's Rule only works on quotients, the forms \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). A product like \(x \ln x\) as \(x \to 0^+\) has the form \(0 \cdot \infty\), so it has to be rewritten as a fraction first. There are always two ways to do that:

Method	What you do	Pattern	Resulting form
Method 1 (blue path)	Flip the factor that goes to 0 into the denominator	\(\dfrac{\infty}{1/0}\)	\(\dfrac{\infty}{\infty}\)
Method 2 (orange path)	Flip the factor that goes to \(\infty\) into the denominator	\(\dfrac{0}{1/\infty}\)	\(\dfrac{0}{0}\)

Both conversions are algebraically correct. But only one of them usually leads somewhere useful. The other one makes the derivatives messier and can leave you with a new \(0 \cdot \infty\) that is worse than the one you started with.

The MicroSim has three examples:

Original form	Method 1	Method 2	Better choice
\(x \ln x\) as \(x \to 0^+\)	\(\dfrac{\ln x}{1/x}\)	\(\dfrac{x}{1/\ln x}\)	Method 1, limit \(= 0\)
\(x e^{-x}\) as \(x \to \infty\)	\(\dfrac{x}{e^x}\)	\(\dfrac{e^{-x}}{1/x}\)	Method 1, limit \(= 0\)
\(x \sin(1/x)\) as \(x \to \infty\)	\(\dfrac{x}{\csc(1/x)}\)	\(\dfrac{\sin(1/x)}{1/x}\)	Method 2, limit \(= 1\)

The third example is there on purpose. If you memorize "always use Method 1," it will burn you.

Choose the conversion that makes the derivative simpler!

Before you flip a factor, ask what its reciprocal's derivative looks like. \(\frac{d}{dx}\left[\frac{1}{x}\right]\) is easy. \(\frac{d}{dx}\left[\frac{1}{\ln x}\right]\) is not.

Delta Moment

"Zero says 'I win, the answer is nothing!' Infinity says 'I win, the answer is everything!' I say: put one of you downstairs in a fraction and let the derivatives sort it out."

How to Use

The MicroSim opens on Example 1, \(x \ln x\), with both methods already worked out. This is the same limit as Example 7 in the chapter. Compare the green card with the red card.
Click the x e⁻ˣ example button. The cards are now hidden.
Look at the colored chips in the top box. The blue chip is the factor that goes to 0 (the one Method 1 flips). The orange chip is the factor that goes to \(\infty\) (the one Method 2 flips).
Predict which factor you would flip into the denominator.
Click Try Method 1 or Try Method 2 (or click the card itself) to see the rewritten quotient and the result of one round of L'Hospital's Rule.
A green band means the derivatives got simpler and the limit can be finished. A red band means the conversion made things worse.
Try the other method too, then repeat with x sin(1/x).
Click Reset to hide the cards and try again.

Iframe Embed Code

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Lesson Plan

Grade Level

Grades 11-12 (college placement Calculus)

Duration

10-15 minutes

Learning Objectives

After using this MicroSim, students will be able to:

Convert an indeterminate product \(0 \cdot \infty\) into a quotient suitable for L'Hospital's Rule (Bloom Level 3)
Identify which factor approaches 0 and which approaches \(\infty\)
Choose the conversion that produces simpler derivatives
Recognize when a conversion has made the limit harder and switch methods

Prerequisites

L'Hospital's Rule for the forms \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\)
Derivatives of \(\ln x\), \(e^x\), and the trigonometric functions
The chain rule

Activities

Study the worked example (3 min): With Example 1 showing, students explain to a partner why the Method 2 card is red. What form does \(-x(\ln x)^2\) have as \(x \to 0^+\)?
Predict and test (5 min): For Example 2, students write down their prediction, try it, then try the other method. They record the form each method produced and the result after one round of L'Hospital's Rule.
Break the pattern (4 min): For Example 3, most students will pick Method 1 because it worked twice. Let them. Then discuss why Method 2 wins this time.
Write the rule (3 min): Students write their own one-sentence rule for choosing which factor to flip.

Discussion Questions

Both conversions are correct algebra. So what makes one of them "wrong" in practice?
In Example 2, why does flipping \(e^{-x}\) cost nothing, while flipping \(x\) makes the problem worse?
In Example 3, what is special about having \(1/x\) inside the sine and in the denominator?
How would you convert \(\lim_{x \to 0^+} \sin x \cdot \ln x\)? Which factor would you flip?

Assessment

Students have met the objective when they can take a new \(0 \cdot \infty\) limit and:

state which factor approaches 0 and which approaches \(\infty\)
write both possible quotients and name the form of each
pick the conversion with the simpler derivatives and finish the limit