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L'Hopital's Rule

  • Use for indeterminate forms: limits that equal
  • Using L’Hôpital’s Rule:
    • Check conditions
    • Find
    • Evaluate to get

L'Hopital's Rule

Given limx → π [cosx/3x], can L'Hopital's Rule be applied?
  • Identify conditions
  • f(π) = − 1
  • g(π) = 3π
It cannot because f(π) ≠ g(π) ≠ 0
Given limx → π [x/sinx], can L'Hopital's Rule be applied?
  • Identify conditions
  • f(0) = 0
  • g(0) = 0
  • f′(0) = 1
  • g′(0) = cos0
  • = 1
Yes, L'Hopital's Rule can be applied.
Given limx → 0 [(√{8 − x} − 8)/x], can L'Hopital's Rule be applied? If so, what is the limit?
  • Identify conditions
  • f(0) = √{8 − 0} − 8
  • = 2√2 − 8
It cannot be applied because f(0) ≠ (0) ≠ 0
Given limθ→ π [(tanθ)/(3cosθ+ 3)], can L'Hopital's Rule be applied? If so, what is the limit?
  • Identify conditions
  • g′(π) = − 3sinπ
  • = 0
It cannot be applied because g′(π) = 0
Given limx → ∞ [(x + 4)/(2 − 5x)], can L'Hopital's Rule be applied? If so, what is the limit?
  • Identify conditions
  • f(∞) = ∞+ 4
  • = ∞
  • g(∞) = 2 − 5(∞)
  • = ∞
  • f′(x) = 1
  • g′(x) = − 5
  • Apply L'Hopital's Rule
limx → ∞ [(x + 4)/(2 − 5x)] = [1/( − 5)]
Given limy → − 2 [(7y + 14)/(3y2 − 12)], can L'Hopital's Rule be applied? If so, what is the limit?
  • Identify conditions
  • f( − 2) = 7y + 14
  • = 7( − 2) + 14
  • = 0
  • g( − 2) = 3y2 − 12
  • = 3( − 2)2 − 12
  • = 3(4) − 12
  • = 0
  • f′( − 2) = 7
  • g′( − 2) = 6( − 2)
  • = − 12
Apply L'Hopital's Rule
limy − 2 [(7y + 14)/(3y2 − 12)] = [7/(−12)]
Given limb → ∞ [(b3 + 8b2)/3], can L'Hopital's Rule be applied? If so, what is the limit?
  • Identify conditions
  • g(∞) = 3
It cannot be applied because g(∞) ≠ 0 or ∞
Find limx → ∞ [(3x)/(9x + 2)]
  • Identify conditions
  • f(∞) = 3
  • = ∞
  • g(∞) = 9x + 2
  • = ∞
  • f′(∞) = (ln3)3
  • g′(x) = 9
  • Apply L'Hopital's Rule
limx → ∞ [(3x)/(9x + 2)] = [((ln3)3 )/9]
Find limx → 0 [(6x2)/(ex − 1)]
  • Identify conditions
  • f(0) = 6x2
  • = 6(9)2
  • = 0
  • g(0) = ex − 1
  • = e0 − 1
  • = 1 − 1
  • = 0
  • f′(0) = 12(0)
  • = 0
  • g′(0) = e0
  • = 1
Apply L'Hopital's Rule
limy → − 2 [(6x2)/(ex − 1)] = [0/1] = 0
Given limx → ∞ [(3x1/3)/(ex(x2 + 1))], can L'Hopital's Rule be applied? If so, what is the limit?
  • Identify conditions
  • f(∞) = 3(∞)1/3
  • = ∞
  • g(∞) = e(∞2 + 1)
  • = ∞
  • f′(∞) = x − 2/3
  • = ∞ − 2/3
  • = 0
  • g′(∞) = ex(2x) + ex(x2)
  • = e(2(∞)) + e(∞2)
  • = ∞
  • Apply L'Hopital's Rule
  • limx∞ [(3x1/3)/(ex(x2 + 1))] = [0/(∞)]
= 0

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

L'Hopital's Rule

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • L'Hopitals Rule 0:09
    • Conditions
    • Limit
  • Example 1 1:21
  • Example 2 2:38
  • Example 3 4:09
  • Example 4 5:01