Difference between revisions of "009A Sample Midterm 1, Problem 2"
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| '''(c)''' <math>1</math> | | '''(c)''' <math>1</math> | ||
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| − | | '''(d)''' <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: -1px">x=1</math> since <math style="vertical-align: -12px">\lim_{x\rightarrow 1}f(x)=f(1).</math> | + | | '''(d)''' <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: -1px">x=1</math> since <math style="vertical-align: -12px">\lim_{x\rightarrow 1}f(x)=f(1).</math> |
|} | |} | ||
[[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 11:20, 18 March 2017
Consider the following function
(a) Find
(b) Find
(c) Find
(d) Is continuous at Briefly explain.
| Foundations: |
|---|
| 1. If |
| then |
| 2. is continuous at if |
Solution:
(a)
| Step 1: |
|---|
| Notice that we are calculating a left hand limit. |
| Thus, we are looking at values of that are smaller than |
| Using the definition of we have |
| Step 2: |
|---|
| Now, we have |
|
|
(b)
| Step 1: |
|---|
| Notice that we are calculating a right hand limit. |
| Thus, we are looking at values of that are bigger than |
| Using the definition of we have |
| Step 2: |
|---|
| Now, we have |
|
|
(c)
| Step 1: |
|---|
| From (a) and (b), we have |
| and |
| Step 2: |
|---|
| Since |
| we have |
(d)
| Step 1: |
|---|
| From (c), we have |
| Also, |
| Step 2: |
|---|
| Since |
| is continuous at |
| Final Answer: |
|---|
| (a) |
| (b) |
| (c) |
| (d) is continuous at since |