Difference between revisions of "022 Exam 1 Sample A, Problem 3"

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|where&thinsp; <math style="vertical-align: 0%">0^{-}</math> can be thought of as "really small negative numbers approaching zero."  Since the handed limits do not agree, the limit as x approaches 5 does not exist.
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|where&thinsp; <math style="vertical-align: 0%">0^{-}</math> can be thought of as "really small negative numbers approaching zero."  Since the handed limits do not agree, the limit as <math style="vertical-align: 0%">x</math> approaches 5 does not exist.
 
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Latest revision as of 11:06, 12 April 2015

Problem 3. Given a function  ,

(a) Find the intervals where is continuous.
(b). Find .
Foundations:  
A function is continuous at a point if
This can be viewed as saying the left and right hand limits exist, and are equal to the value of at .

 Solution:

(a):  
Note that
In order to be continuous at a point , must exist. However, attempting to plug in results in division by zero. Therefore, in interval notation, we have that is continuous on
(b):  
Note that in order for the limit to exist, the limit from both the left and the right must be equal. But
while
where  can be thought of as "really small negative numbers approaching zero." Since the handed limits do not agree, the limit as approaches 5 does not exist.
Final Answer:  
(a): is continuous on
(b): does not exist.

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