March 15th, 2017, 04:56 AM  #1 
Newbie Joined: Mar 2017 From: Norway Posts: 4 Thanks: 0  Limit of x > +1
Hi, I am a new member in here. My daughter has asked me this question and after I solved it she was not agree with my solution. Can anyone help me with the correct solution please? Here is the problem: If (3x² + ax)=a+2 lim x > +1 Find values of "a" And here is what I did: [3(1.1)²+ a(1.1)]= a+2 a+2<=3.6+a.a<a+3 a+2<=3.6+a.a 3.6+a.a<a+3 a>= 16 a< 6 
March 15th, 2017, 06:27 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,784 Thanks: 2198 Math Focus: Mainly analysis and algebra 
I can't make sense of the question, but if there is a limit to be obtained as $x \to 1$ then making $a$ the subject of the equation and finding the limit as $x \to 1$ would seem sensible. But the limit exists only if the $2$ in the question should be a $3$, in which case I get $a=6$.
Last edited by v8archie; March 15th, 2017 at 06:36 AM. 
March 15th, 2017, 07:26 AM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,493 Thanks: 889 Math Focus: Elementary mathematics and beyond 
First of all, $3x^2+ax$ is a polynomial so there's no reason that direct substitution cannot be used. That gives us $3+a=a+2$, which is absurd. I'm assuming $x$ and $a$ are real.

March 15th, 2017, 09:24 AM  #4  
Newbie Joined: Mar 2017 From: Norway Posts: 4 Thanks: 0  Quote:
I really appreciate it if you could show me how to get $a=6$ and why that is the only value for $a$.  
March 15th, 2017, 10:09 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,784 Thanks: 2198 Math Focus: Mainly analysis and algebra 
Perhaps you could write down the full question first.

March 15th, 2017, 10:20 AM  #6 
Newbie Joined: Mar 2017 From: Norway Posts: 4 Thanks: 0  
March 15th, 2017, 11:21 AM  #7 
Senior Member Joined: May 2016 From: USA Posts: 632 Thanks: 257  
March 15th, 2017, 11:25 AM  #8 
Newbie Joined: Mar 2017 From: Norway Posts: 4 Thanks: 0  
March 15th, 2017, 02:46 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,784 Thanks: 2198 Math Focus: Mainly analysis and algebra 
If, as I suggested above the question were $3x^2 + ax=a+3$, this can be rearranged (when $x\ne 1$) to get $$a=\frac{3x^23}{x1}=\frac{3(x+1)\cancel{(x1)}}{\cancel{(x1)}}=3(x+1)$$ Taking the limit of this expression as $x \to 1$ gives $a=6$ 

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