Left hand and right hand limits

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Homework Help Overview

The discussion revolves around determining the left-hand and right-hand limits of the function 1/x^2 - 4 at its vertical asymptotes, specifically at x = 2 and x = -2. The original poster expresses confusion regarding the behavior of the left-hand limit at x = 2, as indicated by their graph and the textbook's assertion of negative infinity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to reconcile their graphical representation with the textbook's claim about the left-hand limit at x = 2. Some participants question the interpretation of the graph in relation to the limit, while others suggest focusing on the behavior of the curve just left of the asymptote.

Discussion Status

Participants are actively engaging with the original poster's confusion, providing insights about the interpretation of limits in relation to the graph. There is a recognition of the importance of accurately representing the function to avoid misunderstandings, but no explicit consensus has been reached regarding the original poster's interpretation.

Contextual Notes

There is a mention of potential confusion arising from the notation of the function, with suggestions to clarify the expression by using parentheses. The original poster's understanding of the limits is challenged by the textbook's information, leading to a deeper exploration of the topic.

grace77
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Problem statement
What is the left hand and right hand limit of 1/x^2 -4 at its vertical asymptote?

Revelant equations
None

Attempt at a solution
It's vertical asymptote are 2 and -2.
ImageUploadedByPhysics Forums1391863762.126560.jpg


I have attached my work . I understand it however at the back of the book it says the left hand limit at 2 is negative infinity. That I don't understand as if you look at the graph I have drawn the left hand limit at 2 is going in a positive direction.

Does anyone have any insight on this? Thank you!
 
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your graph is correct. but your written answer does not match the graph. The left-hand limit at 2 should be where the curve is going, when x is just left of 2. So when the curve is just before x=2, where is the curve going towards: -infinity or +infinity?
 
BruceW said:
your graph is correct. but your written answer does not match the graph. The left-hand limit at 2 should be where the curve is going, when x is just left of 2. So when the curve is just before x=2, where is the curve going towards: -infinity or +infinity?
Oh so you look at it before the curve : in this case it is heading downwards therefore it's a negative infinity
 
yep. you look at the curve just left of the x value you are interested in. p.s. it's better to write the equation as 1/(x^2-4) instead of 1/x^2 -4 since this is confusing without the bracket. You might even lose marks if you hand it in without the bracket. I'm sure I've done that in the past. Although, on paper it is more obvious, since you can put it all under the division sign.
 
BruceW said:
yep. you look at the curve just left of the x value you are interested in. p.s. it's better to write the equation as 1/(x^2-4) instead of 1/x^2 -4 since this is confusing without the bracket. You might even lose marks if you hand it in without the bracket. I'm sure I've done that in the past. Although, on paper it is more obvious, since you can put it all under the division sign.
Ok thank you Bruce! You are a lifesaver!
 
haha, glad to help. But I'll get a big ego if you say things like that!
 

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