Difference between revisions of "009A Sample Midterm 1, Problem 2"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
| Line 41: | Line 41: | ||
|Notice that we are calculating a left hand limit. | |Notice that we are calculating a left hand limit. | ||
|- | |- | ||
| − | |Thus, we are looking at values of <math>x</math> that are smaller than <math>1.</math> | + | |Thus, we are looking at values of <math style="vertical-align: 0px">x</math> that are smaller than <math style="vertical-align: -2px">1.</math> |
|- | |- | ||
| − | |Using the definition of <math>f(x)</math>, we have | + | |Using the definition of <math style="vertical-align: -5px">f(x)</math>, we have |
|- | |- | ||
| <math>\lim_{x\rightarrow 1^-} f(x)=\lim_{x\rightarrow 1^-} x^2.</math> | | <math>\lim_{x\rightarrow 1^-} f(x)=\lim_{x\rightarrow 1^-} x^2.</math> | ||
| Line 71: | Line 71: | ||
|Notice that we are calculating a right hand limit. | |Notice that we are calculating a right hand limit. | ||
|- | |- | ||
| − | |Thus, we are looking at values of <math>x</math> that are bigger than <math>1.</math> | + | |Thus, we are looking at values of <math style="vertical-align: 0px">x</math> that are bigger than <math style="vertical-align: -2px">1.</math> |
|- | |- | ||
| − | |Using the definition of <math>f(x)</math>, we have | + | |Using the definition of <math style="vertical-align: -5px">f(x)</math>, we have |
|- | |- | ||
| <math>\lim_{x\rightarrow 1^+} f(x)=\lim_{x\rightarrow 1^+} \sqrt{x}.</math> | | <math>\lim_{x\rightarrow 1^+} f(x)=\lim_{x\rightarrow 1^+} \sqrt{x}.</math> | ||
| Line 140: | Line 140: | ||
| <math>\lim_{x\rightarrow 1}f(x)=f(1),</math> | | <math>\lim_{x\rightarrow 1}f(x)=f(1),</math> | ||
|- | |- | ||
| − | |<math>f(x)</math> is continuous at <math>x=1.</math> | + | |<math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: -1px">x=1.</math> |
|- | |- | ||
| | | | ||
| Line 155: | Line 155: | ||
| '''(c)''' <math>1</math> | | '''(c)''' <math>1</math> | ||
|- | |- | ||
| − | | '''(d)''' <math>f(x)</math> is continuous at <math>x=1</math> since <math>\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 16:46, 18 February 2017
Consider the following function
(a) Find
(b) Find
(c) Find
(d) Is continuous at Briefly explain.
| Foundations: |
|---|
| 1. If |
| then |
| 2. Definition of continuous |
| 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 |