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 21: | Line 21: | ||
!Foundations: | !Foundations: | ||
|- | |- | ||
− | |'''1.''' If <math style="vertical-align: -15px">\lim_{x\rightarrow a^-} f(x)=\lim_{x\rightarrow a^+} f(x)=c,</math> | + | |'''1.''' If <math style="vertical-align: -15px">\lim_{x\rightarrow a^-} f(x)=\lim_{x\rightarrow a^+} f(x)=c,</math> |
|- | |- | ||
− | | then <math style="vertical-align: -12px">\lim_{x\rightarrow a} f(x)=c.</math> | + | | then <math style="vertical-align: -12px">\lim_{x\rightarrow a} f(x)=c.</math> |
|- | |- | ||
− | |'''2.''' | + | |'''2.''' <math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: 0px">x=a</math> if |
− | |||
− | |||
|- | |- | ||
| <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math> | | <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math> | ||
Line 41: | Line 39: | ||
|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 style="vertical-align: 0px">x</math> that are smaller than <math style="vertical-align: -2px">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 style="vertical-align: -5px">f(x)</math> | + | |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 69: | ||
|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 style="vertical-align: 0px">x</math> that are bigger than <math style="vertical-align: -2px">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 style="vertical-align: -5px">f(x)</math> | + | |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 138: | ||
| <math>\lim_{x\rightarrow 1}f(x)=f(1),</math> | | <math>\lim_{x\rightarrow 1}f(x)=f(1),</math> | ||
|- | |- | ||
− | |<math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: -1px">x=1.</math> | + | |<math style="vertical-align: -5px">f(x)</math> is continuous at <math style="vertical-align: -1px">x=1.</math> |
|- | |- | ||
| | | |
Revision as of 16:46, 26 February 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 |