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Indeterminate forms and L'Hospital's Rule
- 1. Sikkim Institute ofScience & Technology Mathematics Presentation By – Bijay Sharma 1st year - Computer Science & Engineering Department
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- 3. Contents ✘ What dowe mean by indeterminate forms? ✘ Forms of Indeterminates ✘ Why L’Hosipital’s Rule? ✘ Understanding L’Hospital’s Rule ✘ Example Problem using L’Hospital’s Rule ✘ 4 Problems with L’Hospital’s Rule 3
- 4. What do wemean by Indeterminate form? Something is said to be of indeterminate when it has no fixed numeric value. Example: 0 0 , ∞ ∞ , ∞ − ∞, 0°, 1∞ , ∞0 , 0 × ∞
- 5. Big concept Infinity (∞)is not a number, rather, it exists only as an abstract concept. 5
- 6. 6 8 2 = 4 ∴ 𝐼𝑡𝑖𝑠 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 0 0 = 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 ∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 ∞ ∞ = 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 ∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 NB : ∞ ∞ = 1 ∞ − ∞ = 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 ∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 00 = 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 ∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 1∞ = 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 ∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 ∞0 = 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 ∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 0 × ∞ = 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 ∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 Indeterminate forms
- 7. Why do weneed L’Hopital’s Rule ? Let’s say we want to evaluate the given limit lim 𝑥→0 sin 𝑥 𝑥 = 0 0 This limit is indeterminate 7
- 8. L’Hopital Guillaume François Antoine, Marquisde l'Hôpital 8
- 9. L’Hospital’s Rule L'Hospital's ruleuses derivatives to help evaluate limits involving indeterminate forms. Application of the rule often converts an indeterminate form to an expression that can be evaluated by substitution, allowing easier evaluation of the limit.
- 10. L’Hospital’s Rule Suppose thatwe have one of the following cases: lim 𝑥→𝑎 𝑓(𝑥) 𝑔(𝑥) = 0 0 𝑜𝑟 lim 𝑥→𝑎 𝑓(𝑥) 𝑔(𝑥) = ±∞ ±∞ 𝑤ℎ𝑒𝑟𝑒 𝑎 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟, 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦 𝑜𝑟 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦. 𝐼𝑛 𝑡ℎ𝑒𝑠𝑒 𝑐𝑎𝑠𝑒 𝑤𝑒 ℎ𝑎𝑣𝑒, lim 𝑥→𝑎 𝑓(𝑥) 𝑔 𝑥 = lim 𝑥→𝑎 𝑓′(𝑥) 𝑔′(𝑥) So, L’Hospital’s Rule tells us that if we have an indeterminate form or all we need to do is differentiate the numerator and differentiate the denominator and then take the limit. 10
- 11. Example 11 Let’s say wewant to find the solution for 𝐥𝐢𝐦 𝒙→𝟎 𝐬𝐢𝐧 𝒙 𝒙 lim 𝑥→0 sin 𝑥 𝑥 = 0 0 (but, this limit is indeterminate) ∴ 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑎𝑝𝑝𝑙𝑦 𝐿′ 𝐻𝑜𝑠𝑝𝑖𝑡𝑎𝑙′ 𝑠 𝑟𝑢𝑙𝑒 lim 𝑥→0 sin 𝑥 𝑥 = lim 𝑥→0 cos 𝑥 𝑥 = 1 1 = 1
- 12.
- 13. Problem 1 :Evaluate the given limit : lim 𝑡→1 5𝑡4−4𝑡2−1 10−𝑡−9𝑡³ lim 𝑡→1 5𝑡4 − 4𝑡2 − 1 10 − 𝑡 − 9𝑡3 = 5(1)4 − 4(1)2 −1 10 − 1 − 9(1)³ = 0 0 We get an indeterminate form. Hence we will apply L’Hospital’s rule to evaluate the limit ⇒ lim 𝑡→1 5𝑡4 − 4𝑡2 − 1 10 − 𝑡 − 9𝑡³ = lim 𝑡→1 20𝑡3 − 8𝑡 −1 − 27𝑡² = − 3 7 13
- 14. Problem 2 :Evaluate the given limit : lim 𝑥→∞ 𝑒 𝑥 𝑥³ lim 𝑥→∞ 𝑒 𝑥 𝑥² = ∞ ∞ (indeterminate) ∴ 𝑤𝑒 𝑎𝑝𝑝𝑙𝑦 𝐿′ 𝐻𝑜𝑠𝑝𝑖𝑡𝑎𝑙′ 𝑠 𝑟𝑢𝑙𝑒 lim 𝑥→∞ 𝑒 𝑥 𝑥² = lim 𝑥→∞ 𝑒 𝑥 2𝑥 (The new limit also turns out to be indeterminate) We know how to deal with these kinds of limits, We will just apply L’Hospital’s rule again lim 𝑥→∞ 𝑒 𝑥 𝑥² = lim 𝑥→∞ 𝑒 𝑥 2𝑥 = 𝑒 𝑥 2 = ∞ NB – We can apply L’Hospital’s rule more than once to get the results. 14
- 15. Problem 3 :Evaluate the given limit : lim 𝑥→0 𝑥+sin 𝑥 𝑥+cos 𝑥 lim 𝑥→0 𝑥+sin 𝑥 𝑥+cos 𝑥 = 0 1 = 0 Using the L’Hospital’s rule here would have given us the wrong answer. The rule only works on indeterminate forms. 15
- 16. Big concept lim 𝑥→0 𝑓 𝑥. 𝑔(𝑥) = (0)(∞) lim 𝑥→0 𝑓 𝑥 . 𝑔(𝑥) = lim 𝑥→0 𝑓(𝑥) 1 𝑔(𝑥) = lim 𝑥→0 𝑔(𝑥) 1 𝑓(𝑥) Now we can apply L’Hospital’s Rule 16
- 17. Problem 4 :Evaluate the given limit : lim 𝑥→0+ 𝑥. ln 𝑥 lim 𝑥→0+ 𝑥. ln 𝑥 = (0)(∞) (indeterminate) = lim 𝑥→0+ ln 𝑥 𝑥−1 Now lets take the derivative = lim 𝑥→0+ 1 𝑥 −𝑥−2 = lim 𝑥→0+ 1 𝑥 . −𝑥² 1 = lim 𝑥→0+ − 𝑥 = 0 17
- 18. Problem 5 :Evaluate the given limit : lim 𝑥→∞ 4𝑥2−5𝑥 1−3𝑥² lim 𝑥→∞ 4𝑥2−5𝑥 1−3𝑥² = ∞ −∞ (indeterminate) We will apply L’Hospital’s rule. = lim 𝑥→∞ 8𝑥 − 5 −6𝑥 = ∞ −∞ NB – We can apply L’Hospital’s rule more than once. ∴ 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑎𝑝𝑝𝑙𝑦 𝐿′ 𝐻𝑜𝑠𝑝𝑖𝑡𝑎𝑙′ 𝑠 𝑟𝑢𝑙𝑒 𝑎𝑔𝑎𝑖𝑛 = lim 𝑥→∞ 8 −6 = 𝟒 −𝟑 Bonus – Let’s see another way to solve the same problem in the next slide.18
- 19. Bonus : Evaluatethe given limit : lim 𝑥→∞ 4𝑥2−5𝑥 1−3𝑥² lim 𝑥→∞ 4𝑥2−5𝑥 1−3𝑥² = ∞ −∞ (indeterminate) = lim 𝑥→∞ 𝑥²(4 − 5 𝑥) 𝑥²(1 𝑥2 − 3) = lim 𝑥→∞ 4 − 5 𝑥 1 𝑥² − 3 = 𝟒 −𝟑 19
- 20.