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Thread: Weird limits problem

  1. #1
    Junior Member Coco87's Avatar
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    Weird limits problem

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

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

    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!
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  2. #2
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    Quote Originally Posted by Coco87 View Post
    Hey,
    I just got my test back today, and missed a few problems. Two of these problems were similar to this one:

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

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

    first factor

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

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

    $\displaystyle (x^2+a/6x+(a+2)/6) = (x+3)(x-b)
    $

    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 $\displaystyle -\frac{5}{3}$

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

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

    Bobak
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  3. #3
    o_O
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    $\displaystyle \frac{6x^{2} + ax + a + 2}{x^{2} + x - 6} = \frac{6x^2 + ax + a+2}{(x+3)(x-2)}$

    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

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

    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: $\displaystyle 56 - 2a$

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

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

    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.
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  4. #4
    Junior Member Coco87's Avatar
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    Quote Originally Posted by bobak View Post
    $\displaystyle \lim_{x \to -3} \frac{6x^2+ax+a+2}{x^2+x-6}$

    first factor

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

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

    $\displaystyle (x^2+a/6x+(a+2)/6) = (x+3)(x-b)
    $

    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 $\displaystyle -\frac{5}{3}$

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

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

    Bobak
    Hmm... you lost me around $\displaystyle (x^2+a/6x+(a+2)/6) = (x+3)(x-b)
    $

    Quote Originally Posted by o_O View Post
    $\displaystyle \frac{6x^{2} + ax + a + 2}{x^{2} + x - 6} = \frac{6x^2 + ax + a+2}{(x+3)(x-2)}$

    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

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

    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: $\displaystyle 56 - 2a$

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

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

    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!
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  5. #5
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    Allow me to give you a simple solution.
    You want $\displaystyle (x+3)$ to be a factor of $\displaystyle 6x^2 + ax + a + 2$.
    Thus use the factor/root theorem and solve $\displaystyle 6\left( { - 3} \right)^2 + a\left( { - 3} \right) + a + 2 = 0$ for a.
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  6. #6
    Junior Member Coco87's Avatar
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    Quote Originally Posted by Plato View Post
    Allow me to give you a simple solution.
    You want $\displaystyle (x+3)$ to be a factor of $\displaystyle 6x^2 + ax + a + 2$.
    Thus use the factor/root theorem and solve $\displaystyle 6\left( { - 3} \right)^2 + a\left( { - 3} \right) + a + 2 = 0$ for a.
    Thank you good sir! Makes sense now
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