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 Post subject: Limit help pleasePosted: Tue, 13 Sep 2005 00:11:07 UTC
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Joined: Tue, 13 Sep 2005 00:02:28 UTC
Posts: 79
find the following limits. give a graphical representation for each

1) Find lim(x->4) abs(x-4)/x^2-16

Work: abs(x-4)/(x+4)(x-4) = 1/(x+4)=1/8. Is this correct? abs=absolute value. Also, the graphical representation would be a hole in the graph right?

2) lim(x-> pi/4) tan(2x)= tan(pi/2)=dne. correct?

3) lim(x->3) 3-sqrt(x+6)/x-3.

work: multiply top and bottom by 3+sqrt(x+6)

top= 9-(x+6)
bottom: I am having trouble doing the bottom

4) lim(x->1) (1/x -1)/x^2-1= 1-x/x/(x+1)(x-1) =-(x+1)/x=-2 graphical representation: hole in graph

thank you!

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 Post subject: Posted: Tue, 13 Sep 2005 00:42:47 UTC
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3) top=3-x. What's the problem?

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 Post subject: Re: Limit help pleasePosted: Tue, 13 Sep 2005 00:48:17 UTC
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Joined: Tue, 10 Feb 2004 23:54:30 UTC
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This should get you started...

engima89 wrote:
find the following limits. give a graphical representation for each

1) Find lim(x->4) abs(x-4)/x^2-16

Work: abs(x-4)/(x+4)(x-4) = 1/(x+4)=1/8. Is this correct? abs=absolute value. Also, the graphical representation would be a hole in the graph right?

Look at what happens for x>4.

And for x<4,

Do you see where this is going?

Quote:
2) lim(x-> pi/4) tan(2x)= tan(pi/2)=dne. correct?

And what do you know about ?

Oh and for number 3:

-Kwyjibo

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 Post subject: Posted: Tue, 13 Sep 2005 01:15:42 UTC
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Joined: Tue, 13 Sep 2005 00:02:28 UTC
Posts: 79
for 1, that means the limit does not exist, however what would be the graphical representation.

also i dont understand what u mean by what happens at tan(pi/2). i think it is undefined there.

For three, i dont understand how u did the algebraic manipulation.

thanks

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 Post subject: Posted: Tue, 13 Sep 2005 02:16:00 UTC
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Joined: Tue, 10 Feb 2004 23:54:30 UTC
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Location: The Colbert Nation
1. There is a discontinuity at x=4 (what kind?) Draw a graph if you like.

2. Yes, tan(pi/2) is undefined. So the limit doesn't exist. You don't say lim x-> pi/4 (tan(2x)) = tan(pi/2).

3. It's the difference of squares again. I thought it might be a little easier for you.

-Kwyjibo

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 Post subject: Posted: Tue, 13 Sep 2005 18:01:23 UTC
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Joined: Tue, 13 Sep 2005 00:02:28 UTC
Posts: 79
no one commented on number 4, did i do it correctly?

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 Post subject: Posted: Tue, 13 Sep 2005 18:24:39 UTC
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Joined: Mon, 10 Nov 2003 22:38:41 UTC
Posts: 1108

Last edited by stapel on Tue, 18 Oct 2005 16:49:08 UTC, edited 2 times in total.

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 Post subject: Posted: Tue, 13 Sep 2005 18:28:52 UTC
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Joined: Tue, 13 Sep 2005 00:02:28 UTC
Posts: 79
sorry aboout that

lim ( (1/(x) - 1)/x^2-1). note: x^2-1 is divided by the whole numerator.
(x->1)

hope that is a bit more clear

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 Post subject: Posted: Tue, 13 Sep 2005 19:43:27 UTC
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Joined: Sun, 22 Jun 2003 18:00:06 UTC
Posts: 3550
Location: North TX
lim{x->1} [(1/x) - 1]/(x^2 - 1) =

lim{x->1} [(1-x)/x]/[(x+1)(x-1)] =

lim{x->1} (1-x)/[x(x+1)(x-1)] =

lim{x->1} -1/[x(x+1)] = -1/2

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