Author 
Topic 

markmil2002
Average Member
USA
11 Posts 
Posted  09/11/2007 : 22:16:40

I'm trying to find the limit of ((2x9) + 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? 


tkhunny
Advanced Member
USA
1001 Posts 
Posted  09/11/2007 : 22:47:39

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? 


markmil2002
Average Member
USA
11 Posts 
Posted  09/12/2007 : 07:03:20

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 


markmil2002
Average Member
USA
11 Posts 
Posted  09/12/2007 : 07:09:35

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


Subhotosh Khan
Advanced Member
USA
9116 Posts 
Posted  09/12/2007 : 07:34:21

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 (2x9) does not exist when x < 4.5
Thus limit at x > 0 does not make sense 



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