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Really hard limit problem

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
Is there a number "a" such that

limit of ((3x^2 + ax + a + 3) / (x^2 + x -2)) as x approaches -2.

I have attached a picture of this problem for those who need a visual aid and a side for "hints". I have been trying to solve this question with limited success.

Limit problem.JPG
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
You can directly follow the hint

\(\displaystyle \lim_{x \to -2}3x^2+ax+a+3=0\)
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
I realize that, but I don't understand the hints leading up to that point. Can you explain or elaborate? Like the (3x + k).
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Let us look at an easier example

\(\displaystyle \lim_{x \to 2}\frac{x^2-a}{x-2}\)

So , what should be the value of $a$ so that the limit exists and why ?
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
(x-2), right?
 

eddybob123

Active member
Aug 18, 2013
76
I realize that, but I don't understand the hints leading up to that point. Can you explain or elaborate? Like the (3x + k).
Hint #1: It is easy to see that the denominator factors as $(x-1)(x+2)$. Since the limit goes to $x=-2$, we must get rid of the $(x+2)$ term in the denominator. To do this, the numerator also has to factor with an $(x+2)$ term.

Hint #2: The leading coefficient of the numerator is 3, and one of its factors ($(x+2)$) has a leading coefficient of 1, so the other factor has a leading coefficient of 3, so it is in the form $(3x+k)$, for some constant k.

Hint #4: Expanding $(x+2)(3x+k)$ gives $3x^2+(6+k)x+2k$, which in turn gives $6+k=a$ and $2k=a+3$. Using this, it is easy to solve for a and k, which will give the factors of the numerator.

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(x-2), right?
Actually, I'm pretty sure he meant a constant term, so no. Try to find an $a$ such that the numerator contains an $(x-2)$ term.
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
Hint #2: The leading coefficient of the numerator is 3, and one of its factors ($(x+2)$) has a leading coefficient of 1, so the other factor has a leading coefficient of 3, so it is in the form $(3x+k)$, for some constant k.
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Actually, I'm pretty sure he meant a constant term, so no. Try to find an $a$ such that the numerator contains an $(x-2)$ term.
The problem lies on Hint two; I lost you there D:
Actually, I'm pretty sure he meant a constant term, so no. Try to find an $a$ such that the numerator contains an $(x-2)$ term.
Then a could be 4?
 
Last edited:

eddybob123

Active member
Aug 18, 2013
76
The problem lies on Hint two; I lost you there D:
We want the numerator to factor as $3x^2+ax+a+3=(x+2)(mx+k)$ for some coefficients m and k. Expanding this gives $3x^2+ax+a+3=mx^2+kx+2mx+2k$. Therefore, $m=3$ and restrictions to $a$ can be made.

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Then a could be 4?
Yes, it is. The numerator can factor as \(\displaystyle (x+2)(x-2)\)

*Edit* Actually, you don't even need to find restrictions to a. You can solve for it directly using Hint #4.
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
We want the numerator to factor as $3x^2+ax+a+3=(x+2)(mx+k)$ for some coefficients m and k. Expanding this gives $3x^2+ax+a+3=mx^2+kx+2mx+2k$. Therefore, $m=3$ and restrictions to $a$ can be made.

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Yes, it is. The numerator can factor as \(\displaystyle (x+2)(x-2)\)

*Edit* Actually, you don't even need to find restrictions to a. You can solve for it directly using Hint #4.
Thanks, I get it now, but, what led you choose "(mx+k)"? Could you not just have done (x+2)(x+k)?

EDIT: Oh, could "m" be representing any arbitrary number?
 

eddybob123

Active member
Aug 18, 2013
76
Thanks, I get it now, but, what led you choose "(mx+k)"? Could you not just have done (x+2)(x+k)?
Expanding $(x+2)(x+k)$ gives $x^2+(k+2)x+2k$, where the polynomial we want is actually $3x^2+ax+a+3$. Note that the coefficient of $x^2$ in the cases are different. If we want the equality to be satisfied for all x, we want the coefficients to be the same.

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Oh, could "m" be representing any arbitrary number?
It is a variable which we solved for.
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
Expanding $(x+2)(x+k)$ gives $x^2+(k+2)x+2k$, where the polynomial we want is actually $3x^2+ax+a+3$. Note that the coefficient of $x^2$ in the cases are different. If we want the equality to be satisfied for all x, we want the coefficients to be the same.

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It is a variable which we solved for.
If I wanted it in the form $3x^2+ax+a+3$, and I had to multiple (something)(x+2) to obtain it in that form, it wasn't intuitive for me to multiple that by "(mx+K)". It works out, but I wouldn't have came up with that immediately ... so would I have to use like trial and error to do so, or is there some sort of principle?
 

eddybob123

Active member
Aug 18, 2013
76
It works out, but I wouldn't have came up with that immediately ... so would I have to use like trial and error to do so, or is there some sort of principle?
What's wrong with \(\displaystyle (mx+k)\)?
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
What's wrong with \(\displaystyle (mx+k)\)?
Nothing, it's just that I wouldn't have come up with that straight off the bat. If I needed it in the form 3x2+ax+a+3, multiplying (x-2) by "mx+k" wouldn't have been apparent. How were you able to determine so quickly that "mx+k" was the ideal candidate?
 

eddybob123

Active member
Aug 18, 2013
76
How were you able to determine so quickly that "mx+k" was the ideal candidate?
The polynomial you were trying to factor was of degree 2, so there are two factors of degree 1, that is, of the form $(ax+b)(cx+d)$.
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
The polynomial you were trying to factor was of degree 2, so there are two factors of degree 1, that is, of the form $(ax+b)(cx+d)$.
Thank you!