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October 6th, 2008, 11:13 PM   #1
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limit evaluation

Hi,

Given is: Lim x->-2 (x+2)/(x^3 +

I make: Lim x->-2 (x+2)/ [(x+2) (x^2 -2x+4)]

Why is the answer 1/12 ?

Does (x+2)/[(x+2) (...)] simply sets a 1 on the numerator and leaves what's left as denominator? If so, what is the law(s)/method(s) applied to reach this?

I hope some one can help me. Thanks.
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October 7th, 2008, 04:49 AM   #2
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Re: limit evaluation

Hi

One of the rules of limits is that as long as both and exist. In your case, you can

write and I hope you can see why and that giving the required answer of 1/12.
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October 7th, 2008, 08:25 AM   #3
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Re: limit evaluation

Quote:
Originally Posted by sj@vanbunningen.eu
Hi,

Given is: Lim x->-2 (x+2)/(x^3 +

I make: Lim x->-2 (x+2)/ [(x+2) (x^2 -2x+4)]

Why is the answer 1/12 ?
Because as x-> -2, x is never equal to -2, so you can legitimately divide (x+2)/(x+2) to have 1/(x^2 -2x+4)

You now get the result by direct substitution.
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October 20th, 2008, 12:43 AM   #4
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Re: limit evaluation

Quote:
Originally Posted by mattpi
Hi

One of the rules of limits is that as long as both and exist. In your case, you can

write and I hope you can see why and that giving the required answer of 1/12.
I understand the way you solve the problem although there are several points I do not completely get;

as the given formula is: lim x->-2 (x+2)/(x^3 + I would rather make out of lim x->-2 f(x)= lim x->-2 (x+2) and x->-2 g(x)= lim x->-2 (x^3 +

As I rationalize the denominator (that's probably not the right term in this case, but I mean making [(x+2)(x^2 -2x +4)] out of (x^3 +) this also leaves me with (x+2)/[(x+2)(x^2 -2x +4)]= 1/(x^2 -2x +4) which is the same as the quoted formula g(x). here I don't use the rule lim x->a [f(x)g(x)]= lim x->a f(x) lim x->a g(x)

So; I do understand the solution but not the way to get there.
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October 20th, 2008, 07:24 AM   #5
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Re: limit evaluation

Hi

The reason I did not use and is because the limit does not exist.

What I am actually doing is being very pedantic and saying that you can't cancel the factors of x+2; i.e., since does not hold for all x (it is undefined for x=-2). However, we can say that It's a fine distinction, but it sometimes pays to be pedantic!
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October 20th, 2008, 07:34 AM   #6
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Re: limit evaluation

EDIT: Mattpi beat me to it.

Such a tempest in a teapot!

(x+2)/[(x+2)(x^2 -2x + 4)] = 1/(x^2 - 2x + 4) from simple algebraic principles studies well before the introduction to calculus. The working principle from there is that any number [quantity] divided by itself is equal to 1 except for 0/0 which is undefined in a finite number system. In the calculus it is that as x approaches -2, (x+2) is NOT equal to zero, and so the division can be done in a finite number system, those factors being eliminated.

The problem here then is lack of understanding of the algebra, then of the calculus in that direct substitution in the final statement gives the required result. This manner of elimination of factors then substitution is so common in these problems it must have been encountered already.
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