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March 15th, 2017, 04:56 AM   #1
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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
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March 15th, 2017, 06:27 AM   #2
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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.
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March 15th, 2017, 07:26 AM   #3
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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.
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March 15th, 2017, 09:24 AM   #4
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Quote:
Originally Posted by v8archie View Post
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 $a=-6$.
Thank you very much for your explanation.
I really appreciate it if you could show me how to get $a=-6$ and why that is the only value for $a$.
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March 15th, 2017, 10:09 AM   #5
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Perhaps you could write down the full question first.
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March 15th, 2017, 10:20 AM   #6
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Originally Posted by v8archie View Post
Perhaps you could write down the full question first.
This is the exact wording of the problem:

If (3x² + ax)=a+2
lim x --> +1

Find values of "a"
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March 15th, 2017, 11:21 AM   #7
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Quote:
Originally Posted by Farzin View Post
This is the exact wording of the problem:

If (3x² + ax)=a+2
lim x --> +1

Find values of "a"
If that is the problem, it contains one or more typographical errors. There is no such number. Full stop.
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March 15th, 2017, 11:25 AM   #8
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If that is the problem, it contains one or more typographical errors. There is no such number. Full stop.
Thank you very much.
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March 15th, 2017, 02:46 PM   #9
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If, as I suggested above the question were $3x^2 + ax=a+3$, this can be rearranged (when $x\ne 1$) to get $$a=-\frac{3x^2-3}{x-1}=-\frac{3(x+1)\cancel{(x-1)}}{\cancel{(x-1)}}=-3(x+1)$$

Taking the limit of this expression as $x \to 1$ gives $a=-6$
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