My Math Forum limit evaluation

 Calculus Calculus Math Forum

 October 6th, 2008, 10:13 PM #1 Newbie   Joined: Sep 2008 Posts: 9 Thanks: 0 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.
 October 7th, 2008, 03:49 AM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: limit evaluation Hi One of the rules of limits is that $\lim_{x\mapsto a}\left[f(x)g(x)\right]=\lim_{x\mapsto a} f(x)\lim_{x\mapsto a} g(x),$ as long as both $\lim_{x\mapsto a} f(x)$ and $\lim_{x\mapsto a} g(x)$ exist. In your case, you can write $f(x)=\frac{x+2}{x+2}$ and $g(x)=\frac1{x^2-2x+4}.$ I hope you can see why $\lim_{x\mapsto-2}f(x)=1$ and that $\lim_{x\mapsto-2}g(x)=g(-2),$ giving the required answer of 1/12.
October 7th, 2008, 07:25 AM   #3
Senior Member

Joined: Jul 2008

Posts: 895
Thanks: 0

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.

October 19th, 2008, 11:43 PM   #4
Newbie

Joined: Sep 2008

Posts: 9
Thanks: 0

Re: limit evaluation

Quote:
 Originally Posted by mattpi Hi One of the rules of limits is that $\lim_{x\mapsto a}\left[f(x)g(x)\right]=\lim_{x\mapsto a} f(x)\lim_{x\mapsto a} g(x),$ as long as both $\lim_{x\mapsto a} f(x)$ and $\lim_{x\mapsto a} g(x)$ exist. In your case, you can write $f(x)=\frac{x+2}{x+2}$ and $g(x)=\frac1{x^2-2x+4}.$ I hope you can see why $\lim_{x\mapsto-2}f(x)=1$ and that $\lim_{x\mapsto-2}g(x)=g(-2),$ 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.

 October 20th, 2008, 06:24 AM #5 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: limit evaluation Hi The reason I did not use $f(x)=x+2$ and $g(x)=\frac{1}{x^3-8}$ is because the limit $\lim_{x\mapsto-2}g(x)$ 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., $\frac{x+2}{(x+2)(x^2-2x-4)}\neq\frac{1}{x^2-2x-4},$ since $\frac{x+2}{x+2}=1$ does not hold for all x (it is undefined for x=-2). However, we can say that $\lim_{x\mapsto-2}\frac{x+2}{x+2}=1.$ It's a fine distinction, but it sometimes pays to be pedantic!
 October 20th, 2008, 06:34 AM #6 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 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.

 Tags evaluation, limit

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Rejjy Calculus 11 December 25th, 2016 08:56 PM guru123 Calculus 4 October 6th, 2013 02:32 PM mathbalarka Calculus 5 February 18th, 2013 12:53 AM date Calculus 3 June 12th, 2012 11:51 AM notnaeem Calculus 1 January 30th, 2011 12:34 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top