Answer Limit of (x-1)/(x^2)(x+2) as x → 0

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression (x-1) / (x^2)(x+2) as x approaches 0. Participants are exploring the behavior of the numerator and denominator as x approaches this value.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Some participants question whether the limit approaches zero, while others suggest that the limit may not exist due to the behavior of the numerator and denominator. There is mention of L'Hopital's rule and its applicability to this limit.

Discussion Status

The discussion is active, with participants providing different perspectives on the limit's value. Some express uncertainty about the application of L'Hopital's rule, while others clarify that the limit does not exist based on the behavior of the function as x approaches 0.

Contextual Notes

There is a lack of consensus on the interpretation of the limit, with some participants suggesting it approaches -∞ while others state it does not exist. The original poster's understanding of L'Hopital's rule is also noted as a point of confusion.

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lim (x-1) / (x^2)(x+2) as x approach to 0

Is the answer equal to zero??

No matter what mathod I use ,the answer I got are all the same.
 
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the numerator tends to -1 the denominator to zero, doesn't that tell you something? It seems quite simple so perhaps you copied the question down wrongly
 
This is the right question. Nothing is wrong with it.
 
Is the answer does not exist?
 
do you know L'Hopital's rule that is
[tex]\lim_{g(x) \rightarrow 0} \frac{f(x)}{g(x)} = \lim_{g(x) \rightarrow 0} \frac{f'(x)}{g'(x)}[/tex]
if you use that here you should get zero
also mathematica gives me zero for the answer as well
 
Sorry, I don`t know about L'Hopital's rule.
 
L'Hopital's rule says that if g and f both have limit 0 as a goes to a, then
[tex]lim_{x\rightarrow a}\frac{f(x)}{g(x)}= \frac{lim_{x\rightarrow a}f(x)}{lim_{x\rightarrow a}g(x)}[/tex]
Since, in this problem, the limit of the numerator, [itex]lim_{x\rightarrow 0}x-1[/itex] is -1, not 0, LHopital's rule does not apply.
As x approaches 0, the numerator stays around -1 while the denominator goes to 0: the fraction goes toward [itex]-\infty[/itex].
Some people would say "the limit does not exist". Others would say the limit is [itex]-\infty[/itex] which is just a way of saying the limit does not exist in a particular way.
 
Thanks everyone.
 

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