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Math Help - finding 2 limits. Unsure of answer.

  1. #1
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    finding 2 limits. Unsure of answer.

    I'm trying to find limits for 2 situations.

    1. lim x->a

    f(x)-f(a) if f(x)=x^2+4 gave me lim x ->a = 2a
    --------
    x-a

    2. lim x->a

    f(x)*(x^3-a^3) if f(x)=a^3 and g(x)=a^2 so I got lim x->a = 2a^2(a+1)
    ---------------
    (x-a)*a^2

    My main concern is that so far in every situation where we had a letter instead of a number we could get rid of the letters eventually and get a number.

    I'm wondering if the answers are correct and if not what might be the correct answer and/or steps to get there.

    Thank you.
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  2. #2
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    Re: finding 2 limits. Unsure of answer.

    Quote Originally Posted by Raitor View Post

    1. lim x->a

    f(x)-f(a) if f(x)=x^2+4 gave me lim x ->a = 2a
    --------
    x-a
    What is f(a) if f(x) = x^2 + 4?

    To start you out:
    \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = \lim_{x \to a} \frac{x^2 - a^2}{x - a}

    The rest I leave up to you.

    -Dan
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  3. #3
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    Re: finding 2 limits. Unsure of answer.

    [QUOTE=Raitor;735446]I'm trying to find limits for 2 situations.

    1. lim x->a

    f(x)-f(a) if f(x)=x^2+4 gave me lim x ->a = 2a
    --------
    x-a
    Unfortunately, The internet does not always "respect" spaces. f(x)= (x^2+ 4)/(x- a). But since we cannot divide by 0, f(a) does not exist so f(x)- f(a) does not exist. Do you mean just "lim f(x) as x goes to a?. Well that also does not exist.

    I think you mean "find \lim_{x\to a}\frac{f(x)- f(a)}{x- a} where f(x)= x^2+ 4.
    Okay, if that is true, f(x)- f(a)= x^2+ 4- (a^2+ 4)= x^2- a^2= (x- a)(x+ a).

    2. lim x->a

    f(x)*(x^3-a^3) if f(x)=a^3 and g(x)=a^2 so I got lim x->a = 2a^2(a+1)
    ---------------
    (x-a)*a^2
    What does "g(x)= x^2" have to do with this? There is no "g" in what you give.
    If you mean \lim_{x\to a}\frac{a^3(x^3- a^3}{(x- a)(a^2)} then you need to know that x^3- a^3= (x- a)(x^2+ ax+ a^2). So this is the same as \lim_{x\to a}\frac{a^3/a^2}\frac{x- a}{x- a}(x^2+ ax+ a^2)

    My main concern is that so far in every situation where we had a letter instead of a number we could get rid of the letters eventually and get a number.
    No, the answer is NOT a number, it will be a function of a.

    I'm wondering if the answers are correct and if not what might be the correct answer and/or steps to get there.

    Thank you.
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  4. #4
    Forum Admin topsquark's Avatar
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    Re: finding 2 limits. Unsure of answer.

    Quote Originally Posted by Raitor View Post
    2. lim x->a

    f(x)*(x^3-a^3) if f(x)=a^3 and g(x)=a^2 so I got lim x->a = 2a^2(a+1)
    ---------------
    (x-a)*a^2
    If f(x) = a^3 then f(x) is a constant. Same for g(x). I think you need to re-state this.

    -Dan
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  5. #5
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    Re: finding 2 limits. Unsure of answer.

    Thank you for your quick answers.
    I'll rewrite the whole thing also taking into account what you told me and I made a couple mistakes while writing down the second problem.

    1. Evaluate \lim_{x\to a}\frac{f(x)- f(a)}{x-a} where f(x)=x^2+4

    I ended up doing this :

    \lim_{x\to a}\frac{f(x)- f(a)}{x-a} = \lim_{x\to a}\frac{x^2-a^2}{x-a} = \lim_{x\to a}\frac{(x+a)(x-a)}{x-a} = \lim_{x\to a}(x+a) = (a+a) = 2a

    2. For \lim_{x\to a}f(x) = a^3 and \lim_{x\to a}g(x) = a^2 evaluate the limit of :

    \lim_{x\to a}\frac{f(x)*(x^3-a^3)}{(x-a)*g(x)}

    For this I did the following.

    \lim_{x\to a}\frac{f(x)*(x^3-a^3)}{(x-a)*g(x)} = \lim_{x\to a}\frac{a^3*(x^3-a^3)}{(x-a)*a^2} = \lim_{x\to a}\frac{a^3*(x-a)(x^2+ax+a^2)}{(x-a)*a^2} = \lim_{x\to a}\frac{a^3*(x^2+ax+a^2)}{a^2} = \lim_{x\to a}a(x^2+ax+a^2) = a(a^2+2a+a^2) = 2a^2(a+1)


    Wish I had known about this way to insert equations on the forums the first time around.

    Would those answers be correct? What about the way I wrote the equations?

    Thank you again!
    Last edited by Raitor; September 11th 2012 at 11:09 AM.
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