L'Hospital's Rule (College Board AP® Calculus AB)

Study Guide

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Indeterminate forms

What is an indeterminate form?

  • An indeterminate form is a mathematical expression of one of the two following forms:

    • 0 over 0

    • fraction numerator plus-or-minus infinity over denominator plus-or-minus infinity end fraction

  • The value of an indeterminate form is undefined

    • Dividing by 0 always gives an undefined expression

    • And note that, for example, infinity over infinity is not equal to 1

      • infinity is not a number

      • so it can't be canceled to simplify a fraction

  • Sometimes attempting to evaluate a limit using substitution leads to one of the indeterminate forms given above

    • L'Hospital's rule provides a method for dealing with limits of that form

Examiner Tips and Tricks

Other limit methods will also sometimes work when substitution gives an indeterminate form

  • For example algebraic simplification, multiplying by conjugates or multiplying by reciprocals

    • See the 'Evaluating Limits Analytically' study guide

Evaluating limits using L'Hospital's rule

What is L'Hospital’s Rule?

  • L'Hospital's rule (sometimes written as L’Hôpital’s rule) is a method for finding the value of certain limits using calculus

    • Specifically, it allows us to attempt to evaluate the limit of a quotient fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction 

    • for which attempting to evaluate the limit by substitution returns one of the indeterminate forms 0 over 0 or fraction numerator plus-or-minus infinity over denominator plus-or-minus infinity end fraction

  • For such a quotient function, L'Hospital's rule says that

    • limit as x rightwards arrow a of fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction equals limit as x rightwards arrow a of fraction numerator f to the power of apostrophe open parentheses x close parentheses over denominator g to the power of apostrophe open parentheses x close parentheses end fraction

    • I.e., you can take the derivatives of the numerator and denominator

      • and attempt to evaluate the limit again in that form

How do I evaluate a limit using L’Hospital’s Rule?

  • STEP 1
    Check that the limit of the quotient results in one of the indeterminate forms given above

    • I.e., check that limit as x rightwards arrow a of fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction equals fraction numerator f left parenthesis a right parenthesis over denominator g left parenthesis a right parenthesis end fraction equals 0 over 0 or fraction numerator plus-or-minus infinity over denominator plus-or-minus infinity end fraction

  • STEP 2
    Find the derivatives of the numerator and denominator of the quotient

  • STEP 3
    Check whether the limit limit as x rightwards arrow a of fraction numerator f to the power of apostrophe left parenthesis x right parenthesis over denominator g to the power of apostrophe left parenthesis x right parenthesis end fraction exists

  • STEP 4
    If that limit does exist, then limit as x rightwards arrow a of fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction equals limit as x rightwards arrow a of fraction numerator f to the power of apostrophe left parenthesis x right parenthesis over denominator g to the power of apostrophe left parenthesis x right parenthesis end fraction

  • STEP 5
    If limit as x rightwards arrow a of fraction numerator f to the power of apostrophe left parenthesis x right parenthesis over denominator g to the power of apostrophe left parenthesis x right parenthesis end fraction equals fraction numerator f to the power of apostrophe left parenthesis a right parenthesis over denominator g to the power of apostrophe left parenthesis a right parenthesis end fraction equals 0 over 0or fraction numerator plus-or-minus infinity over denominator plus-or-minus infinity end fraction then you may repeat the process by considering limit as x rightwards arrow a of fraction numerator f to the power of apostrophe apostrophe end exponent left parenthesis x right parenthesis over denominator g to the power of apostrophe apostrophe end exponent left parenthesis x right parenthesis end fraction (and possibly higher order derivatives after that)

    • As long as the limits continue giving indeterminate forms you may continue applying L’Hospital’s rule

    • Each time this happens find the next set of derivatives and consider the limit again

Examiner Tips and Tricks

Before beginning to use L'Hospital's rule to evaluate a limit

  • Be sure to confirm that using substitution gives an indeterminate form

  • Otherwise L'Hospital's rule is not valid

Worked Example

Use L’Hospital’s rule to evaluate each of the following limits:

(a) limit as x rightwards arrow infinity of fraction numerator 5 x plus 17 over denominator 4 minus 3 x end fraction

Answer:

This limit could also be found by 'multiplying by reciprocals', but the question says to use L'Hospital's rule

First check that substitution gives an indeterminate form, so L'Hospital's rule is valid

limit as x rightwards arrow infinity of fraction numerator 5 x plus 17 over denominator 4 minus 3 x end fraction equals fraction numerator infinity over denominator negative infinity end fraction which is an indeterminate form

Find the derivatives of the numerator and denominator

fraction numerator d over denominator d x end fraction open parentheses 5 x plus 17 close parentheses equals 5

fraction numerator d over denominator d x end fraction open parentheses 4 minus 3 x close parentheses equals negative 3

Apply L'Hospital's Rule, limit as x rightwards arrow a of fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction equals limit as x rightwards arrow a of fraction numerator f to the power of apostrophe left parenthesis x right parenthesis over denominator g to the power of apostrophe left parenthesis x right parenthesis end fraction

limit as x rightwards arrow infinity of fraction numerator 5 x plus 17 over denominator 4 minus 3 x end fraction equals limit as x rightwards arrow infinity of fraction numerator 5 over denominator negative 3 end fraction

There's no x in that final expression, so x going to infinity doesn't matter!

limit as x rightwards arrow infinity of fraction numerator 5 x plus 17 over denominator 4 minus 3 x end fraction equals negative 5 over 3

(b) limit as x rightwards arrow 0 of fraction numerator x cubed over denominator negative 2 x plus sin space 2 x end fraction

Answer:

First check that substitution gives an indeterminate form, so L'Hospital's rule is valid

limit as x rightwards arrow 0 of fraction numerator x cubed over denominator negative 2 x plus sin space 2 x end fraction equals fraction numerator open parentheses 0 close parentheses cubed over denominator negative 2 open parentheses 0 close parentheses plus sin open parentheses 0 close parentheses end fraction equals 0 over 0 which is an indeterminate form

Find the derivatives of the numerator and denominator

fraction numerator d over denominator d x end fraction open parentheses x cubed close parentheses equals 3 x squared

fraction numerator d over denominator d x end fraction open parentheses negative 2 x plus sin 2 x close parentheses equals negative 2 plus 2 cos 2 x

Apply L'Hospital's Rule, limit as x rightwards arrow a of fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction equals limit as x rightwards arrow a of fraction numerator f to the power of apostrophe left parenthesis x right parenthesis over denominator g to the power of apostrophe left parenthesis x right parenthesis end fraction

table row cell limit as x rightwards arrow 0 of fraction numerator x cubed over denominator negative 2 x plus sin space 2 x end fraction end cell equals cell limit as x rightwards arrow 0 of fraction numerator 3 x squared over denominator negative 2 plus 2 cos 2 x end fraction end cell row blank equals cell fraction numerator 3 open parentheses 0 close parentheses squared over denominator negative 2 plus 2 cos open parentheses 0 close parentheses end fraction end cell row blank equals cell fraction numerator 0 over denominator negative 2 plus 2 open parentheses 1 close parentheses end fraction end cell row blank equals cell 0 over 0 end cell end table

That is another indeterminate form, so we can repeat the process again

fraction numerator d over denominator d x end fraction open parentheses 3 x squared close parentheses equals 6 x

fraction numerator d over denominator d x end fraction open parentheses negative 2 plus 2 cos 2 x close parentheses equals negative 4 sin 2 x

Apply L'Hospital's rule again

table row cell limit as x rightwards arrow 0 of fraction numerator 3 x squared over denominator negative 2 plus 2 cos 2 x end fraction end cell equals cell limit as x rightwards arrow 0 of fraction numerator 6 x over denominator negative 4 sin 2 x end fraction end cell row blank equals cell fraction numerator 6 open parentheses 0 close parentheses over denominator negative 4 sin open parentheses 0 close parentheses end fraction end cell row blank equals cell fraction numerator 0 over denominator negative 4 open parentheses 0 close parentheses end fraction end cell row blank equals cell 0 over 0 end cell end table

And that is also an indeterminate form, so we can repeat the process again

fraction numerator d over denominator d x end fraction open parentheses 6 x close parentheses equals 6

fraction numerator d over denominator d x end fraction open parentheses negative 4 sin 2 x close parentheses equals negative 8 cos 2 x

Apply L'Hospital's rule again

table row cell limit as x rightwards arrow 0 of fraction numerator 6 x over denominator negative 4 sin 2 x end fraction end cell equals cell limit as x rightwards arrow 0 of fraction numerator 6 over denominator negative 8 cos 2 x end fraction end cell row blank equals cell fraction numerator 6 over denominator negative 8 cos open parentheses 0 close parentheses end fraction end cell row blank equals cell fraction numerator 6 over denominator negative 8 open parentheses 1 close parentheses end fraction end cell row blank equals cell negative 6 over 8 end cell end table

That is not an indeterminate form, so that's the answer we're looking for

Simplify the fraction

limit as x rightwards arrow 0 of fraction numerator x cubed over denominator negative 2 x plus sin space 2 x end fraction equals negative 3 over 4

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Roger B

Author: Roger B

Expertise: Maths

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.