Problem

Evaluate the limit, if it exists. $ \displayst…

View

Problem 10 Medium Difficulty

(a) What is wrong with the following equation?
$$ \frac {x^2 + x - 6}{x - 2} = x + 3 $$

(b) In view of part (a), explain why the equation
$$ \lim_{x \to 2}\frac {x^2 + x - 6}{x - 2} = \lim_{x \to 2}(x + 3) $$
is correct.

Answer

a. see work for answer
b. $\lim _{x \rightarrow 2} \frac{x^{2}+x-6}{x-2}=\lim _{x \rightarrow 2}(x+3)$

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

Video Transcript

we had the expression X squared plus x minus six over x minus two. We can factor the numerator. The numerator would factor into X plus straight Times X -2. And of course in the denominator, we have X -2. Now we'd like to cancel out these X -2s. Uh but we can do so under one assumption That X is not equal to two. If X was equal to two, then the X -2 that you did have in the denominator before you try to cancel it out would have been zero and you can't have zero in the denominator, you cannot divide by zero. So X is not allowed to equal to As long as X is not allowed to equal to, you can cancel the X -2s and get x plus straight. So it's not as simple as saying that this expression equals X plus three. This expression does equal X-plus three. As long as you add the condition X is not allowed to equal to because remember effects was to the original denominator would have been zero. Can't divide by zero. No. If we want to take the limit of this function as X approaches to Now, if we want to take the limit of this original function as X approaches to uh then we can uh find that limit by taking the limit as X approaches to of the X plus straight. The reason we're allowed to do that is when we find a limit of this function as X approaches to. The limit is only concerned with X approaching to um X does not become two. X is not does not get to equal to when we're finding limits, we're only interested interested in what is happening as X approaches to. So as long as uh since finding the limit as X approaches to, we're not going to be having X equal to two, then we're allowed to have this function written as X plus three because this function was allowed to equal X plus three as long as X was not equal to two. When we're finding the limit as X approaches to, we're not going to have X equaling to only X approaching two. So the limit of this function as X approaches to is simply equal to the limit of this simpler function as X approaches to. And of course this is a nice continuous function. So we can find this limit as X approaches to simply by substituting two and for X. So the limit of X plus three as X approaches to is two plus three, Of course, that's five.

Leon D.

Temple University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 3

Calculating Limits Using the Limit Laws

Problem

Evaluate the limit, if it exists. $ \displayst…

Problem 10 Medium Difficulty

(a) What is wrong with the following equation?
$$ \frac {x^2 + x - 6}{x - 2} = x + 3 $$

(b) In view of part (a), explain why the equation
$$ \lim_{x \to 2}\frac {x^2 + x - 6}{x - 2} = \lim_{x \to 2}(x + 3) $$
is correct.

Answer

a. see work for answer
b. $\lim _{x \rightarrow 2} \frac{x^{2}+x-6}{x-2}=\lim _{x \rightarrow 2}(x+3)$

More Answers

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 1 / AB Educators

Anna Marie V.

Kayleah T.

Kristen K.

Joseph L.

Calculus 1 / AB Courses

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

Watch More Solved Questions in Chapter 2

Video Transcript

Calculus: Early Transcendentals

Anna Marie V.

Caleb E.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

a. see work for answerb. $\lim _{x \rightarrow 2} \frac{x^{2}+x-6}{x-2}=\lim _{x \rightarrow 2}(x+3)$

Calculus: Early Transcendentals

Anna Marie V.

Kayleah T.

Kristen K.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

Anna Marie V.

Caleb E.

Samuel H.

Michael J.

a. see work for answer
b. $\lim _{x \rightarrow 2} \frac{x^{2}+x-6}{x-2}=\lim _{x \rightarrow 2}(x+3)$