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# Lesson 17: Indeterminate Forms and L'Hopital's Rule (Section 041 handout)

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L'Hopital's Rule allows us to resolve limits of indeterminate form: 0/0, infinity/infinity, infinity-infinity, 0^0, 1^infinity, and infinity^0

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### Lesson 17: Indeterminate Forms and L'Hopital's Rule (Section 041 handout)

1. 1. Section 3.7 Indeterminate Forms and L’Hˆopital’s Rule V63.0121.041, Calculus I New York University November 3, 2010 Announcements Announcements V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 2 / 24 Objectives Know when a limit is of indeterminate form: indeterminate quotients: 0/0, ∞/∞ indeterminate products: 0 × ∞ indeterminate diﬀerences: ∞ − ∞ indeterminate powers: 00 , ∞0 , and 1∞ V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 3 / 24 Notes Notes Notes 1 Section 3.7 : L’Hˆopital’s RuleV63.0121.041, Calculus I November 3, 2010
2. 2. Experiments with funny limits lim x→0 sin2 x x = 0 lim x→0 x sin2 x does not exist lim x→0 sin2 x sin(x2) = 1 lim x→0 sin 3x sin x = 3 All of these are of the form 0 0 , and since we can get diﬀerent answers in diﬀerent cases, we say this form is indeterminate. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 4 / 24 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. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 5 / 24 More about dividing limits 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: lim x→0+ 1 x = +∞ lim x→0− cos x x3 = −∞ An exception would be something like lim x→∞ 1 1 x sin x = lim x→∞ x csc x. which does not exist and is not inﬁnite. Even less predictable: numerator and denominator both go to zero. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 6 / 24 Notes Notes Notes 2 Section 3.7 : L’Hˆopital’s RuleV63.0121.041, Calculus I November 3, 2010
3. 3. 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 in each case “is” 0 0 , and therefore nonexistent because this expression is undeﬁned. The limit is of the form 0 0 , which means we cannot evaluate it with our limit laws. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 7 / 24 Indeterminate forms are like Tug Of War Which side wins depends on which side is stronger. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 8 / 24 Outline L’Hˆopital’s Rule Other Indeterminate Limits Indeterminate Products Indeterminate Diﬀerences Indeterminate Powers V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 9 / 24 Notes Notes Notes 3 Section 3.7 : L’Hˆopital’s RuleV63.0121.041, Calculus I November 3, 2010
4. 4. The Linear Case Question If f and g are lines and f (a) = g(a) = 0, what is lim x→a f (x) 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) g(x) = m1 m2 V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 10 / 24 The Linear Case, Illustrated x y y = f (x) y = g(x) a x f (x) g(x) f (x) g(x) = f (x) − f (a) g(x) − g(a) = (f (x) − f (a))/(x − a) (g(x) − g(a))/(x − a) = m1 m2 V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 11 / 24 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? The derivative! V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 12 / 24 Notes Notes Notes 4 Section 3.7 : L’Hˆopital’s RuleV63.0121.041, Calculus I November 3, 2010
5. 5. Theorem of the Day 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 x→a f (x) = 0 and lim x→a g(x) = 0 or lim x→a f (x) = ±∞ and lim x→a g(x) = ±∞ Then lim x→a f (x) g(x) = lim x→a f (x) g (x) , if the limit on the right-hand side is ﬁnite, ∞, or −∞. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 13 / 24 Meet the Mathematician: L’Hˆopital 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 Fran¸cois Antoine, Marquis de L’Hˆopital (French, 1661–1704) V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 14 / 24 Revisiting the previous examples Example lim x→0 sin2 x x H = lim x→0 2 sin x sin x → 0 cos x 1 = 0 Example lim x→0 sin2 x numerator → 0 sin x2 denominator → 0 H = lim x→0 ¡2 sin x cos x numerator → 0 (cos x2) (¡2x denominator → 0 ) H = lim x→0 cos2 x − sin2 x numerator → 1 cos x2 − 2x2 sin(x2) denominator → 1 = 1 Example lim x→0 sin 3x sin x H = lim x→0 3 cos 3x cos x = 3. V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 15 / 24 Notes Notes Notes 5 Section 3.7 : L’Hˆopital’s RuleV63.0121.041, Calculus I November 3, 2010
6. 6. Beware of Red Herrings Example Find lim x→0 x cos x Solution V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 16 / 24 Outline L’Hˆopital’s Rule Other Indeterminate Limits Indeterminate Products Indeterminate Diﬀerences Indeterminate Powers V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 17 / 24 Indeterminate products Example Find lim x→0+ √ x ln x This limit is of the form 0 · (−∞). Solution Jury-rig the expression to make an indeterminate quotient. Then apply L’Hˆopital’s Rule: lim x→0+ √ x ln x = lim x→0+ ln x 1/ √ x H = lim x→0+ x−1 −1 2x−3/2 = lim x→0+ −2 √ x = 0 V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 18 / 24 Notes Notes Notes 6 Section 3.7 : L’Hˆopital’s RuleV63.0121.041, Calculus I November 3, 2010
7. 7. Indeterminate diﬀerences Example lim x→0+ 1 x − cot 2x This limit is of the form ∞ − ∞. Solution V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 19 / 24 Indeterminate powers Example Find lim x→0+ (1 − 2x)1/x Take the logarithm: ln lim x→0+ (1 − 2x)1/x = lim x→0+ ln (1 − 2x)1/x = lim x→0+ ln(1 − 2x) x This limit is of the form 0 0 , so we can use L’Hˆopital: lim x→0+ ln(1 − 2x) x H = lim x→0+ −2 1−2x 1 = −2 This is not the answer, it’s the log of the answer! So the answer we want is e−2 . V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 21 / 24 Another indeterminate power limit Example lim x→0 (3x)4x Solution V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 22 / 24 Notes Notes Notes 7 Section 3.7 : L’Hˆopital’s RuleV63.0121.041, Calculus I November 3, 2010
8. 8. Summary Form Method 0 0 L’Hˆopital’s rule directly ∞ ∞ L’Hˆopital’s rule directly 0 · ∞ jiggle to make 0 0 or ∞ ∞ . ∞ − ∞ factor to make an indeterminate product 00 take ln to make an indeterminate product ∞0 ditto 1∞ ditto V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 23 / 24 Final Thoughts L’Hˆopital’s Rule only works on indeterminate quotients Luckily, most indeterminate limits can be transformed into indeterminate quotients L’Hˆopital’s Rule gives wrong answers for non-indeterminate limits! V63.0121.041, Calculus I (NYU) Section 3.7 L’Hˆopital’s Rule November 3, 2010 24 / 24 Notes Notes Notes 8 Section 3.7 : L’Hˆopital’s RuleV63.0121.041, Calculus I November 3, 2010