# Really hard limit problem

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#### ZaidAlyafey

##### Well-known member
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You can directly follow the hint

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

#### Rido12

##### Well-known member
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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
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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 ?

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(x-2), right?

#### eddybob123

##### Active member
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
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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?

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#### eddybob123

##### Active member
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
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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
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
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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
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
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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
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
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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!