University:
Whitman College
Course:
MATH-124 | Introduction to Calculus
Academic year:

2022
Views:

45

Pages:

2
Author:

Blake Dillon

Worksheet, Section 2.8 (l’Hospital’s Rule) Name: Date: Directions: You are encouraged to work in groups, but you don’t need to do so. We’ll turn in what we could finish at the end of class. Find each limit, using l’Hospital’s rule if appropriate. x3 − 2x2 + 1 x→1 x3 − 1 SOLUTION: This is 0/0 form, so we can use l’Hospital’s rule: 1. lim x3 − 2x2 + 1 3x2 + 4x 7 = lim = 3 2 x→1 x→1 x −1 3x 3 lim 2. lim x→0 sin(4x) tan(5x) SOLUTION: This is 0/0 form, so we can use l’Hospital’s rule: lim x→0 4 cos(4x) 4 sin(4x) = lim = tan(5x) x→0 5 sec2 (5x) 5 (For the last step, recall that cos(0) = 1 and sec(0) = 1) e2t − 1 t→0 sin(t) 3. lim SOLUTION: This is 0/0 form, so we can use l’Hospital’s rule: 2e2t e2t − 1 = lim =2 t→0 cos(t) t→0 sin(t) lim x2 + x − 6 x→2 x−2 SOLUTION: This is 0/0 form, so we can use l’Hospital’s rule: 4. lim x2 + x − 6 2x + 1 = lim =5 x→2 x→2 x−2 1 lim sin(x) x→π 1 − cos(x) First evaluate at π to see what form we have: 5. lim− sin(π) 0 0 = = =0 1 − cos(π) 1 − (−1) 2 Therefore, the limit is 0 (this function is continuous at π). Side note: If you try to use l’Hospital’s rule on this, you’ll get an incorrect answer, since the fraction is not in the right form... Turn Page Over... ln(x) x x→0 SOLUTION: The numerator, ln(x) → −∞ as x → 0+ , and x → 0, so this is of the form “−∞/0”, which is not the right form for l’Hospital’s rule. Rather, recall that 1/x → ∞ as x → 0+ , so we might think of this as: ln(x) 1 lim+ = lim+ ln(x) · = “ − ∞ · ∞” x x x→0 x→0 6. lim+ so we see that the fraction will still tend to −∞ as x → 0+ . 8 t8 − 1 8t7 = = lim t→1 t5 − 1 t→1 5t4 5 7. lim 8. lim xe−x/2 x→∞ SOLUTION: Here we have a product rather than a fraction, but e−x/2 = fraction out of this: lim xe−x/2 = lim x→∞ x→∞ 1 , ex/2 so we can make a x ex/2 This is now ∞/∞, so we can use l’Hospital’s rule: lim x→∞ 1 x = lim x→∞ (1/2)ex/2 ex/2 The denominator is getting larger and larger while the numerator is constant, therefore overall, this fraction goes to zero. arcsin(x) x SOLUTION: We have 0/0, so apply l’Hospital’s rule: 9. lim x→0 arcsin(x) = lim x→0 x→0 x lim √ 1 1−x2 1 =1 x3x −1 SOLUTION: We have 0/0, so apply l’Hospital’s rule: 10. lim x→0 3x lim x3x 3x + x3x ln(3) 1 = lim = − 1 x→0 3x ln(3) ln(3) x→0 3x ex − 1 − x x→0 x2 SOLUTION: We have 0/0, so apply l’Hospital’s rule: 11. lim ex − 1 − x ex − 1 = lim 2 x→0 x→0 x 2x lim We still have 0/0, so apply l’Hospital’s rule again. ex − 1 ex 1 = lim = x→0 x→0 2 2x 2 lim

Calculus Worksheet, Section 2.8 (l’Hospital’s Rule)

Related Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock