Calculus Troubles: Finding Limits and Understanding 0/0

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    Calculus Limits
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Homework Help Overview

The original poster is working on finding limits in calculus, specifically encountering a 0/0 form after substitution in a limit problem. They express confusion regarding whether this indicates that the limit does not exist and seek clarification on their approach.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of encountering a 0/0 form and whether it suggests the limit does not exist. Questions are raised about the behavior of the function as it approaches a certain point, including considerations of positive infinity and the necessity of limits from both sides being equal.

Discussion Status

The discussion is active with various interpretations being explored regarding the nature of limits that approach infinity and the conditions under which limits exist. Some participants provide insights into the graphical representation of the function to aid understanding.

Contextual Notes

There is an emphasis on the original poster's level of experience in calculus, which may influence their understanding of limits and the implications of different forms encountered during evaluation.

cvc121
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Homework Statement


Find the following limits, if they exist.

Homework Equations

The Attempt at a Solution


I have just started calculus and am having trouble with 3 a). I get 0/0 after substitution so I factored but still get 0 in the denominator. Does this indicate that the limit does not exist? Am I doing the questions correctly? My work is attached below. Thanks! All help is very much appreciated.
View attachment 96417https://www.physicsforums.com/attachments/96417/
 
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You're taking the limit of a constant (3) divided by something which approaches zero. What do you get if you divide a positive constant by an infinitesimal quantity?
 
Undefined = DNE?
 
Or would it be positive infinity?
 
You could also consider the graph of ##\frac{x+2}{x-1}##. It should be clear from that that the limit doesn't exist.
 
cvc121 said:
Or would it be positive infinity?
Yes, but you can approach 1 from both sides. ##lim_{x\rightarrow 1^+} \frac{x+2}{x-1}## must be equal to ##lim_{x\rightarrow 1^-} \frac{x+2}{x-1}## for the limit to exist. Are they?
 
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Saying that a limit "is positive infinity" is just saying that "the limit does not exist" but giving a specific reason- that the values get larger and larger and larger without upper bound, as opposed to getting lower and lower without lower bound or jumping around without getting close to anyone specific number.
 
HallsofIvy said:
Saying that a limit "is positive infinity" is just saying that "the limit does not exist" but giving a specific reason- that the values get larger and larger and larger without upper bound, as opposed to getting lower and lower without lower bound or jumping around without getting close to anyone specific number.

Very nice explanation.
 

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