Difference between revisions of "009A Sample Midterm 1, Problem 2"

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|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
 
|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-} f(x)=\lim_{x\rightarrow 1^-} x^2.</math>  
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-} f(x)=\lim_{x\rightarrow 1^-} x^2.</math>  
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|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
 
|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^+} f(x)=\lim_{x\rightarrow 1^+} \sqrt{x}.</math>  
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^+} f(x)=\lim_{x\rightarrow 1^+} \sqrt{x}.</math>  
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1}f(x)=f(1),</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp; <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>
 
|-
 
|-
 
|
 
|
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|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>1</math>
 
|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>1</math>
 
|-
 
|-
|&nbsp; &nbsp; '''(d)''' &nbsp; &nbsp; <math>f(x)</math> is continuous at <math>x=1</math> since <math>\lim_{x\rightarrow 1}f(x)=f(1)</math>
+
|&nbsp; &nbsp; '''(d)''' &nbsp; &nbsp; <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

Return to Sample Exam