Problem

Find the limit. Use l'Hospital's Rule where appro…

01:00

Problem 40 Hard Difficulty

Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.

$ \lim_{x\to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x} $

Answer

$\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin x}=2$

More Answers

02:05

Wen Z.

Topics

Derivatives

Differentiation

Volume

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Kayleah T.

Harvey Mudd College

Caleb E.

Baylor University

Samuel H.

University of Nottingham

Michael J.

Idaho State University

Watch More Solved Questions in Chapter 4

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Problem 88

Problem 89

Problem 90

Problem 91

Problem 92

Video Transcript

we know this function is in the form zero over zero. Therefore, we know we can use the hospital's rule. We know this is equivalent to zero over zero. What this means is that we apply the hospital's rule second time. The two is now a constant. Therefore it just completely cancels. This is equivalent to zero over zero We apply The happy tells me we've gone. See that now it's e to the X plus feet of the negative acts over co sign acts Plugin We end up with two over one. This gives us an actual answer of Jim.

WZ

Wen Z.

Topics

Derivatives

Differentiation

Volume

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Top Calculus 2 / BC Educators

Catherine R.

Missouri State University

Anna Marie V.

Campbell University

Heather Z.

Oregon State University

Joseph L.

Boston College