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Lesson 18: Indeterminate Forms and L'Hôpital's Rule
 

Lesson 18: Indeterminate Forms and L'Hôpital's Rule

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L'Hôpital's Rule is not a magic bullet (or a sledgehammer) but it does allow us to find limits of indeterminate forms such as 0/0 and ∞/∞. With some algebra we can use it to resolve other ...

L'Hôpital's Rule is not a magic bullet (or a sledgehammer) but it does allow us to find limits of indeterminate forms such as 0/0 and ∞/∞. With some algebra we can use it to resolve other indeterminate forms such as ∞-∞ and 0^0.

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    Lesson 18: Indeterminate Forms and L'Hôpital's Rule Lesson 18: Indeterminate Forms and L'Hôpital's Rule Presentation Transcript

    • Section 3.7 Indeterminate Forms and L’Hôpital’s Rule V63.0121.034, Calculus I November 2, 2009 Announcements No office hours today Spring forward, fall back, yadda yadda yadda . . . . . .
    • Experiments with funny limits sin2 x lim x→0+ x . . . . . . .
    • Experiments with funny limits sin2 x lim =0 x→0+ x . . . . . . .
    • Experiments with funny limits sin2 x lim =0 x→0+ x x lim x→0 sin2 x . . . . . . .
    • Experiments with funny limits sin2 x lim =0 x→0+ x x lim does not exist x→0 sin2 x . . . . . . .
    • Experiments with funny limits sin2 x lim =0 x→0+ x x lim does not exist x→0 sin2 x . sin2 x lim x→0 sin(x2 ) . . . . . .
    • Experiments with funny limits sin2 x lim =0 x→0+ x x lim does not exist x→0 sin2 x . sin2 x lim =1 x→0 sin(x2 ) . . . . . .
    • Experiments with funny limits sin2 x lim =0 x→0+ x x lim does not exist x→0 sin2 x . sin2 x lim =1 x→0 sin(x2 ) sin 3x lim x→0 sin x . . . . . .
    • Experiments with funny limits sin2 x lim =0 x→0+ x x lim does not exist x→0 sin2 x . sin2 x lim =1 x→0 sin(x2 ) sin 3x lim =3 x→0 sin x . . . . . .
    • Experiments with funny limits sin2 x lim =0 x→0+ x x lim does not exist x→0 sin2 x . sin2 x lim =1 x→0 sin(x2 ) sin 3x lim =3 x→0 sin x 0 All of these are of the form , and since we can get different 0 answers in different cases, we say this form is indeterminate. . . . . . .
    • Outline Indeterminate Forms L’Hôpital’s Rule Relative Rates of Growth Application to Indeterminate Products Application to Indeterminate Differences Application to Indeterminate Powers Summary . . . . . .
    • Recall Recall the limit laws from Chapter 2. Limit of a sum is the sum of the limits . . . . . .
    • Recall Recall the limit laws from Chapter 2. Limit of a sum is the sum of the limits Limit of a difference is the difference of the limits . . . . . .
    • Recall Recall the limit laws from Chapter 2. Limit of a sum is the sum of the limits Limit of a difference is the difference of the limits Limit of a product is the product of the limits . . . . . .
    • Recall Recall the limit laws from Chapter 2. Limit of a sum is the sum of the limits Limit of a difference is the difference of the limits Limit of a product is the product of the limits Limit of a quotient is the quotient of the limits ... whoops! This is true as long as you don’t try to divide by zero. . . . . . .
    • We know dividing by zero is bad. Most of the time, if an expression’s numerator approaches a finite number and denominator approaches zero, the quotient approaches some kind of infinity. For example: 1 cos x lim = +∞ lim = −∞ x→0+ x x→0− x3 . . . . . .
    • We know dividing by zero is bad. Most of the time, if an expression’s numerator approaches a finite number and denominator approaches zero, the quotient approaches some kind of infinity. For example: 1 cos x lim = +∞ lim = −∞ x→0+ x x→0− x3 An exception would be something like 1 lim = lim x csc x. x→∞ 1 sin x x→∞ x which doesn’t exist. . . . . . .
    • We know dividing by zero is bad. Most of the time, if an expression’s numerator approaches a finite number and denominator approaches zero, the quotient approaches some kind of infinity. For example: 1 cos x lim = +∞ lim = −∞ x→0+ x x→0− x3 An exception would be something like 1 lim = lim x csc x. x→∞ 1 sin x x→∞ x which doesn’t exist. Even less predictable: numerator and denominator both go to zero. . . . . . .
    • Language Note It depends on what the meaning of the word “is” is Be careful with the language here. We are not saying that the limit 0 in each case “is” , and 0 therefore nonexistent because this expression is undefined. The limit is of the form 0 , which means we 0 cannot evaluate it with our limit laws. . . . . . .
    • Indeterminate forms are like Tug Of War Which side wins depends on which side is stronger. . . . . . .
    • Outline Indeterminate Forms L’Hôpital’s Rule Relative Rates of Growth Application to Indeterminate Products Application to Indeterminate Differences Application to Indeterminate Powers Summary . . . . . .
    • The Linear Case Question If f and g are lines and f(a) = g(a) = 0, what is f(x) lim ? x→a g(x) . . . . . .
    • The Linear Case Question If f and g are lines and f(a) = g(a) = 0, what is f(x) lim ? x→a g(x) Solution The functions f and g can be written in the form f(x) = m1 (x − a) g(x) = m2 (x − a) So f (x ) m1 = g (x ) m2 . . . . . .
    • The Linear Case, Illustrated y . y . = g(x) y . = f(x) g . (x) a . f .(x) . . . x . x . f(x) f(x) − f(a) (f(x) − f(a))/(x − a) m1 = = = g(x) g(x) − g(a) (g(x) − g(a))/(x − a) m2 . . . . . .
    • What then? But what if the functions aren’t linear? . . . . . .
    • What then? But what if the functions aren’t linear? Can we approximate a function near a point with a linear function? . . . . . .
    • What then? But what if the functions aren’t linear? Can we approximate a function near a point with a linear function? What would be the slope of that linear function? . . . . . .
    • Theorem (L’Hopital’s Rule) Suppose f and g are differentiable functions and g′ (x) ̸= 0 near a (except possibly at a). Suppose that lim f(x) = 0 and lim g(x) = 0 x→a x→a or lim f(x) = ±∞ and lim g(x) = ±∞ x→a x→a Then f(x) f′ (x) lim = lim ′ , x→a g(x) x→a g (x) if the limit on the right-hand side is finite, ∞, or −∞. . . . . . .
    • Meet the Mathematician: L’Hôpital wanted to be a military man, but poor eyesight forced him into math did some math on his own (solved the “brachistocrone problem”) paid a stipend to Johann Bernoulli, who proved this theorem and named it after him! Guillaume François Antoine, Marquis de L’Hôpital (1661–1704) . . . . . .
    • Revisiting the previous examples Example sin2 x lim x→0 x . . . . . .
    • Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim x→0 x x→0 1 . . . . . .
    • Revisiting the previous examples Example . in x → 0 s sin2 x H 2 sin x.cos x lim = lim x→0 x x→0 1 . . . . . .
    • Revisiting the previous examples Example . in x → 0 s sin2 x H 2 sin x.cos x lim = lim =0 x→0 x x→0 1 . . . . . .
    • Revisiting the previous examples Example . in x → 0 s sin2 x H 2 sin x.cos x lim = lim =0 x→0 x x→0 1 Example sin2 x lim x→0 sin x2 . . . . . .
    • Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example . umerator → 0 n sin2 x. lim x→0 sin x2 . . . . . .
    • Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example . umerator → 0 n sin2 x. lim . x→0 sin x2 . enominator → 0 d . . . . . .
    • Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example . umerator → 0 n sin2 x. H  sin x cos x 2 lim 2. = lim x→0 sin x x→0 (cos x2 ) (x ) 2 . enominator → 0 d . . . . . .
    • Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example . umerator → 0 n sin2 x H .  sin x cos x 2 lim = lim x→0 sin x2 x→0 (cos x2 ) (x ) 2 . . . . . .
    • Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example . umerator → 0 n sin2 x H .  sin x cos x 2 lim = lim .) x→0 sin x2 x→0 (cos x2 ) (x 2 . enominator → 0 d . . . . . .
    • Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example . umerator → 0 n sin2 x H . cos2 x − sin2 x  sin x cos x H 2 lim = lim lim .) = x→0 cos x2 − 2x2 sin(x2 ) x→0 sin x2 x→0 (cos x2 ) (x 2 . enominator → 0 d . . . . . .
    • Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example . umerator → 1 n sin2 x H  sin x cos x H 2 cos2 x − sin2 x. lim = lim = lim x→0 sin x2 x→0 (cos x2 ) (x ) 2 x→0 cos x2 − 2x2 sin(x2 ) . . . . . .
    • Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example . umerator → 1 n sin2 x H  sin x cos x H 2 cos2 x − sin2 x. lim = lim = lim . x→0 sin x2 x→0 (cos x2 ) (x ) 2 x→0 cos x2 − 2x2 sin(x2 ) . enominator → 1 d . . . . . .
    • Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example sin2 x H  sin x cos x H 2 cos2 x − sin2 x lim = lim = lim =1 x→0 sin x2 x→0 (cos x2 ) (x ) 2 x→0 cos x2 − 2x2 sin(x2 ) . . . . . .
    • Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example sin2 x H  sin x cos x H 2 cos2 x − sin2 x lim = lim = lim =1 x→0 sin x2 x→0 (cos x2 ) (x ) 2 x→0 cos x2 − 2x2 sin(x2 ) Example sin 3x H 3 cos 3x lim = lim = 3. x→0 sin x x→0 cos x . . . . . .
    • Example Find x lim x→0 cos x . . . . . .
    • Beware of Red Herrings Example Find x lim x→0 cos x Solution The limit of the denominator is 1, not 0, so L’Hôpital’s rule does not apply. The limit is 0. . . . . . .
    • Theorem Let r be any positive number. Then ex lim = ∞. x→∞ xr . . . . . .
    • Theorem Let r be any positive number. Then ex lim = ∞. x→∞ xr Proof. If r is a positive integer, then apply L’Hôpital’s rule r times to the fraction. You get ex H H ex lim = . . . = lim = ∞. x→∞ xr x→∞ r! For example, if r = 3, three invocations of L’Hôpital’s Rule give us ex H ex H ex H ex lim = lim = lim = lim =∞ x→∞ x3 x→∞ 3 · x2 x→∞ 2 · 3x x→∞ 1 · 2 · 3 . . . . . .
    • If r is not an integer, let m be the smallest integer greater than r. Then if x > 1, xr < xm , so ex ex > m. xr x The right-hand side tends to ∞ by the previous step. . . . . . .
    • If r is not an integer, let m be the smallest integer greater than r. Then if x > 1, xr < xm , so ex ex > m. xr x The right-hand side tends to ∞ by the previous step. For example, if r = 1/2, r < 1 so for x > 1 ex ex > x1/2 x which gets arbitrarily large. . . . . . .
    • Theorem Let r be any positive number. Then ln x lim = 0. x→∞ xr . . . . . .
    • Theorem Let r be any positive number. Then ln x lim = 0. x→∞ xr Proof. One application of L’Hôpital’s Rule here suffices: ln x H 1 /x 1 lim = lim r−1 = lim r = 0. x→∞ xr x→∞ rx x→∞ rx . . . . . .
    • Indeterminate products Example Find √ lim x ln x x→0+ This limit is of the form 0 · (−∞). . . . . . .
    • Indeterminate products Example Find √ lim x ln x x→0+ This limit is of the form 0 · (−∞). Solution Jury-rig the expression to make an indeterminate quotient. Then apply L’Hôpital’s Rule: √ lim x ln x x→0+ . . . . . .
    • Indeterminate products Example Find √ lim x ln x x→0+ This limit is of the form 0 · (−∞). Solution Jury-rig the expression to make an indeterminate quotient. Then apply L’Hôpital’s Rule: √ ln x lim x ln x = lim √ x→0+ x→0+ 1/ x . . . . . .
    • Indeterminate products Example Find √ lim x ln x x→0+ This limit is of the form 0 · (−∞). Solution Jury-rig the expression to make an indeterminate quotient. Then apply L’Hôpital’s Rule: √ ln x H x −1 lim x ln x = lim √ = lim x→0+ x→0+ 1/ x x→0+ − 1 x−3/2 2 . . . . . .
    • Indeterminate products Example Find √ lim x ln x x→0+ This limit is of the form 0 · (−∞). Solution Jury-rig the expression to make an indeterminate quotient. Then apply L’Hôpital’s Rule: √ ln x H x −1 lim x ln x = lim √ = lim x→0+ x→0+ 1/ x x→0+ − 1 x−3/2 2 √ = lim −2 x x→0+ . . . . . .
    • Indeterminate products Example Find √ lim x ln x x→0+ This limit is of the form 0 · (−∞). Solution Jury-rig the expression to make an indeterminate quotient. Then apply L’Hôpital’s Rule: √ ln x H x −1 lim x ln x = lim √ = lim x→0+ x→0+ 1/ x x→0+ − 1 x−3/2 2 √ = lim −2 x = 0 x→0+ . . . . . .
    • Indeterminate differences Example ( ) 1 lim − cot 2x x→0 x This limit is of the form ∞ − ∞. . . . . . .
    • Indeterminate differences Example ( ) 1 lim − cot 2x x→0 x This limit is of the form ∞ − ∞. Solution Again, rig it to make an indeterminate quotient. sin(2x) − x cos(2x) lim x→0+ x sin(2x) . . . . . .
    • Indeterminate differences Example ( ) 1 lim − cot 2x x→0 x This limit is of the form ∞ − ∞. Solution Again, rig it to make an indeterminate quotient. sin(2x) − x cos(2x) H cos(2x) + 2x sin(2x) lim = lim+ x→0+ x sin(2x) x→0 2x cos(2x) + sin(2x) . . . . . .
    • Indeterminate differences Example ( ) 1 lim − cot 2x x→0 x This limit is of the form ∞ − ∞. Solution Again, rig it to make an indeterminate quotient. sin(2x) − x cos(2x) H cos(2x) + 2x sin(2x) lim = lim+ x→0+ x sin(2x) x→0 2x cos(2x) + sin(2x) =∞ . . . . . .
    • Indeterminate differences Example ( ) 1 lim − cot 2x x→0 x This limit is of the form ∞ − ∞. Solution Again, rig it to make an indeterminate quotient. sin(2x) − x cos(2x) H cos(2x) + 2x sin(2x) lim = lim+ x→0+ x sin(2x) x→0 2x cos(2x) + sin(2x) =∞ The limit is +∞ becuase the numerator tends to 1 while the denominator tends to zero but remains positive. . . . . . .
    • Checking your work . tan 2x lim = 1, so for small x, x→0 2x 1 tan 2x ≈ 2x. So cot 2x ≈ and . 2x 1 1 1 1 − cot 2x ≈ − = →∞ x x 2x 2x as x → 0+ . . . . . . .
    • Indeterminate powers Example Find lim (1 − 2x)1/x x→0+ . . . . . .
    • Indeterminate powers Example Find lim (1 − 2x)1/x x→0+ Take the logarithm: ( ) ( ) 1/x ln(1 − 2x) ln lim (1 − 2x) = lim+ ln (1 − 2x)1/x = lim+ x→0 + x→0 x→0 x . . . . . .
    • Indeterminate powers Example Find lim (1 − 2x)1/x x→0+ Take the logarithm: ( ) ( ) 1/x ln(1 − 2x) ln lim (1 − 2x) = lim+ ln (1 − 2x)1/x = lim+ x→0 + x→0 x→0 x 0 This limit is of the form , so we can use L’Hôpital: 0 −2 ln(1 − 2x) H 1−2x lim = lim+ = −2 x→0+ x x→0 1 . . . . . .
    • Indeterminate powers Example Find lim (1 − 2x)1/x x→0+ Take the logarithm: ( ) ( ) 1/x ln(1 − 2x) ln lim (1 − 2x) = lim+ ln (1 − 2x)1/x = lim+ x→0 + x→0 x→0 x 0 This limit is of the form , so we can use L’Hôpital: 0 −2 ln(1 − 2x) H 1−2x lim = lim+ = −2 x→0+ x x→0 1 This is not the answer, it’s the log of the answer! So the answer we want is e−2 . . . . . . .
    • Example lim (3x)4x x→0 . . . . . .
    • Example lim (3x)4x x→0 Solution ln lim (3x)4x = lim ln(3x)4x = lim 4x ln(3x) x→0+ x→0+ x→0+ ln(3x) H 3/3x = lim+ = lim+ x→0 1/4x x→0 −1/4x2 = lim (−4x) = 0 x→0+ So the answer is e0 = 1. . . . . . .
    • Summary Form Method 0 0 L’Hôpital’s rule directly ∞ ∞ L’Hôpital’s rule directly 0 ∞ 0·∞ jiggle to make 0 or ∞. ∞−∞ factor to make an indeterminate product 00 take ln to make an indeterminate product ∞0 ditto 1∞ ditto . . . . . .
    • Final thoughts L’Hôpital’s Rule only works on indeterminate quotients Luckily, most indeterminate limits can be transformed into indeterminate quotients L’Hôpital’s Rule gives wrong answers for non-indeterminate limits! . . . . . .