Problem

Show that $ f $ is continuous on $ (-\infty, \inf…

View

Question

Answered step-by-step

Problem 38 Easy Difficulty

Use continuity to evaluate the limit.

$ \displaystyle \lim_{x \to 4} 3^{\sqrt{x^2 - 2x - 4}} $

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Daniel Jaimes

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 5

Continuity

Discussion

You must be signed in to discuss.

Top Calculus 1 / AB Educators

Watch More Solved Questions in Chapter 2

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

Video Transcript

This is a problem. Number 38 of the Stuart Calculus eighth Edition, section 2.5. Use continuity to evaluate the limit. The limited expertise for of the function. Three to the power of the square root function, um, of the quantity X squared minus two X minus four And we have our definition of continuity your and operate where if a function is this continues at a point. A. The limit as X approaches sing of dysfunction have is equal to the function, have evaluated a So let's determine the continuity of dysfunction Within the Scrooge function. We have a polynomial which is continuous on all real on the domain, about real numbers, the square root function as a domain associated with all values creator than or equal to zero. So if there's any exercises that make this negative within the square root, those would be not included in the domain for continuity. And then we have the exponential function here, three to this power, our which is continuous for all values of X. So the only real restriction is within the square root. However we see that we have X approaching for for gives us here a value for the polynomial, which is positive, so there will be no domain restrictions. We can be sure that as they purchased for this entire function is certainly continuous, and we can use our definition to evaluate this limit. Three. Race to the square root four. Squared It's two times four mine's for So there's going to be three to the square root 16, minus eight it is for which is three to the power square would afford, which is three squared gives us nine And we saw this limit using our definition of continuity The answer.

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem

Show that $ f $ is continuous on $ (-\infty, \inf…

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 1 / AB Educators

Grace He

Catherine Ross

Kayleah Tsai

Caleb Elmore

Calculus 1 / AB Courses

Recommended Videos

Watch More Solved Questions in Chapter 2

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Kayleah Tsai

Caleb Elmore

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Catherine Ross

Kayleah Tsai

Caleb Elmore

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Kayleah Tsai

Caleb Elmore

James Stewart

Calculus: Early Transcendentals