Weird limits problem

**Coco87** · May 20th 2008, 03:08 PM

Hey,
I just got my test back today, and missed a few problems. Two of these problems were similar to this one:

$\lim_{x \to -3} \frac{6x^2+ax+a+2}{x^2+x-6}$

With this problem, you're trying to find the value of 'a' where it will make the limit possible. The answers are: a=28. the limit: 1.6

I saw this and had no clue how to work it, didn't even see it in the homework

Could any explain?

Thanks!

**bobak** · May 20th 2008, 03:30 PM

Originally Posted by Coco87

Hey,
I just got my test back today, and missed a few problems. Two of these problems were similar to this one:

$\lim_{x \to -3} \frac{6x^2+ax+a+2}{x^2+x-6}$

With this problem, you're trying to find the value of 'a' where it will make the limit possible. The answers are: a=28. the limit: 1.6

I saw this and had no clue how to work it, didn't even see it in the homework

Could any explain?

Thanks!

$\lim_{x \to -3} \frac{6x^2+ax+a+2}{x^2+x-6}$

first factor

$\lim_{x \to -3} \frac{6x^2+ax+a+2}{(x+3)(x-2)}$

What you require is a value of a such that the numerator has a factor of $(x+3)$ so your limit exists. so you numerator must factor in the form of $(x+3)(x-b)$ for some real number "b". take out a factor of 6 to make thing easier.

$(x^2+a/6x+(a+2)/6) = (x+3)(x-b)<br />$

by comparing coefficients you can solve the equation for a and b, the given value of a is correct and for checking purposed only b is $-\frac{5}{3}$

so the limit is reduced to $\lim_{x \to -3} \frac{6(x+5/3)}{(x-2)}$

Did you follow that all okay? Let me know if you want something further explained

Bobak

**o_O** · May 20th 2008, 03:37 PM

$\frac{6x^{2} + ax + a + 2}{x^{2} + x - 6} = \frac{6x^2 + ax + a+2}{(x+3)(x-2)}$

We can see that the (x+3) factor is causing the problem if we directly evaluated the limit by plugging in x = -3. So, let's assume that the numerator is divisible by (x+3) so that

$\frac{(x+3)(\text{stuff})}{(x+3)(x+2)}$

the (x+3)'s will cancel and all that remains is to plug in x = -3 directly.

So, we do long division and you should get the remainder: $56 - 2a$

Since we want (x+3) to factor in nicely, we can assume that the remainder is 0. i.e. $56 - 2a = 0 \quad \Rightarrow a = 28$

So we have: $\frac{6x^{2} + 28x + 30}{(x+3)(x+2)} = \frac{2(x+3)(3x+5)}{(x+3)(x+2)}$

You know where to go from there

--------------------------------------

Or if you learned l'hopital's, you could go that route and set the numerator equal to 0 and solve for a. Then differentiate and solve the limit.

**Coco87** · May 20th 2008, 04:34 PM

Originally Posted by bobak

$\lim_{x \to -3} \frac{6x^2+ax+a+2}{x^2+x-6}$

first factor

$\lim_{x \to -3} \frac{6x^2+ax+a+2}{(x+3)(x-2)}$

What you require is a value of a such that the numerator has a factor of $(x+3)$ so your limit exists. so you numerator must factor in the form of $(x+3)(x-b)$ for some real number "b". take out a factor of 6 to make thing easier.

$(x^2+a/6x+(a+2)/6) = (x+3)(x-b)<br />$

by comparing coefficients you can solve the equation for a and b, the given value of a is correct and for checking purposed only b is $-\frac{5}{3}$

so the limit is reduced to $\lim_{x \to -3} \frac{6(x+5/3)}{(x-2)}$

Did you follow that all okay? Let me know if you want something further explained

Bobak

Hmm... you lost me around $(x^2+a/6x+(a+2)/6) = (x+3)(x-b)<br />$

Originally Posted by o_O

$\frac{6x^{2} + ax + a + 2}{x^{2} + x - 6} = \frac{6x^2 + ax + a+2}{(x+3)(x-2)}$

We can see that the (x+3) factor is causing the problem if we directly evaluated the limit by plugging in x = -3. So, let's assume that the numerator is divisible by (x+3) so that

$\frac{(x+3)(\text{stuff})}{(x+3)(x+2)}$

the (x+3)'s will cancel and all that remains is to plug in x = -3 directly.

So, we do long division and you should get the remainder: $56 - 2a$

Since we want (x+3) to factor in nicely, we can assume that the remainder is 0. i.e. $56 - 2a = 0 \quad \Rightarrow a = 28$

So we have: $\frac{6x^{2} + 28x + 30}{(x+3)(x+2)} = \frac{2(x+3)(3x+5)}{(x+3)(x+2)}$

You know where to go from there

--------------------------------------

Or if you learned l'hopital's, you could go that route and set the numerator equal to 0 and solve for a. Then differentiate and solve the limit.

Ah, I see what you did there. Poor factoring on my part. Though, is long division necessary?

Thank you both for responding!

**Plato** · May 20th 2008, 05:24 PM

Allow me to give you a simple solution.
You want $(x+3)$ to be a factor of $6x^2 + ax + a + 2$ .
Thus use the factor/root theorem and solve $6\left( { - 3} \right)^2 + a\left( { - 3} \right) + a + 2 = 0$ for a.

**Coco87** · May 20th 2008, 06:24 PM

Originally Posted by Plato

Allow me to give you a simple solution.
You want $(x+3)$ to be a factor of $6x^2 + ax + a + 2$ .
Thus use the factor/root theorem and solve $6\left( { - 3} \right)^2 + a\left( { - 3} \right) + a + 2 = 0$ for a.

Thank you good sir! Makes sense now

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