August 7th, 2012, 02:21 AM  #1 
Newbie Joined: Sep 2011 Posts: 23 Thanks: 0  Work check on a trig. limit
I am hoping someone can double check to make sure I didn't do anything illegal (mathematically), especially in step 2 and 3. I am not sure if I just made things up, or if I am allowed to work it this way by some math principles? Thanks! Robert 
August 7th, 2012, 02:27 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs  Re: Work check on a trig. limit
You cannot factor the 3 out of the sine function's argument (even though this coincidentally produces the correct result). I would write: 
August 7th, 2012, 02:43 AM  #3 
Newbie Joined: Sep 2011 Posts: 23 Thanks: 0  Re: Work check on a trig. limit
Do you know why we can not factor out the 3 if it produces the same answer? I have done three problems like this and have got the answer correct. Check this one out: I am wondering why its wrong when I am getting them right? 
August 7th, 2012, 02:53 AM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs  Re: Work check on a trig. limit
The reason I would give is that for , we have: The second problem, I would solve as follows (barring the use of L'Hôpital's rule): 
August 7th, 2012, 02:14 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,433 Thanks: 561  Re: Work check on a trig. limit
(x > 0) Lim sin(kx)/x = klim sin(kx)/kx = k. That is why it works.


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