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 Pre-Calculus and Calculus
 Limits
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markmil2002
Average Member

USA
11 Posts

Posted - 09/11/2007 :  22:16:40  Show Profile
I'm trying to find the limit of ((2x-9) + 3)/x as x approaches 0. I know that with the original equation if x = 3 then it is 0/0. I tried using the reciprical trick and simplified it down to x/(((2x+9)+3), but that still leaves me without a limit. Is there something I am missing?
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tkhunny
Advanced Member

USA
1001 Posts

Posted - 09/11/2007 :  22:47:39  Show Profile
Something fishy on this one. 2(0) - 9 < 0 How is that going to work under a square root and still have a Real limit?
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markmil2002
Average Member

USA
11 Posts

Posted - 09/12/2007 :  07:03:20  Show Profile
No, I made a mistake. Instead of x appraches 0 it should be as x approaches 3. I tested it with an x value of 3 and it gave me 0/0. Sorry about the confusion
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markmil2002
Average Member

USA
11 Posts

Posted - 09/12/2007 :  07:09:35  Show Profile
one last correction. It is as x apporaches 0, with x equaling 0. Sorry for the extra mistake, lol, long days of work.
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Subhotosh Khan
Advanced Member

USA
9117 Posts

Posted - 09/12/2007 :  07:34:21  Show Profile
quote:
Originally posted by markmil2002

one last correction. It is as x apporaches 0, with x equaling 0. Sorry for the extra mistake, lol, long days of work.



The function (2x-9) does not exist when x < 4.5

Thus limit at x -> 0 does not make sense
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