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

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In summary, the limit of (x-1)/(x^2)(x+2) as x approaches 0 is equal to -∞ or does not exist. This can be determined by using L'Hopital's rule, which states that the limit of a fraction can be found by taking the limits of the numerator and denominator separately. However, since the limit of the numerator is -1, which is not 0, L'Hopital's rule does not apply in this case. Therefore, the limit is either -∞ or does not exist, depending on how you interpret it.
  • #1
frozen7
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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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  • #2
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
 
  • #3
This is the right question. Nothing is wrong with it.
 
  • #4
Is the answer does not exist?
 
  • #5
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
 
  • #6
Sorry, I don`t know about L'Hopital's rule.
 
  • #7
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.
 
  • #8
Thanks everyone.
 

What is the limit of (x-1)/(x^2)(x+2) as x approaches 0?

The limit of (x-1)/(x^2)(x+2) as x approaches 0 is undefined.

How do you determine the limit of (x-1)/(x^2)(x+2) as x approaches 0?

To determine the limit of (x-1)/(x^2)(x+2) as x approaches 0, you can use the algebraic limit theorem or plug in values of x that approach 0 from both the left and right sides.

Is the limit of (x-1)/(x^2)(x+2) as x approaches 0 equal to 0?

No, the limit of (x-1)/(x^2)(x+2) as x approaches 0 is undefined, not equal to 0.

What does it mean for a limit to be undefined?

A limit is undefined if the function does not approach a single value as the input approaches the specified value. In this case, as x approaches 0, the function (x-1)/(x^2)(x+2) does not approach a single value, so the limit is undefined.

Can the limit of (x-1)/(x^2)(x+2) as x approaches 0 be evaluated using L'Hopital's rule?

Yes, L'Hopital's rule can be used to evaluate the limit of (x-1)/(x^2)(x+2) as x approaches 0. However, it is important to note that this rule can only be applied if the limit is in an indeterminate form, which in this case, it is not.

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