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Math Help - annoying limit

  1. #1
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    annoying limit

    Hey
    i've just been trying to work out this limit
    its been hours and my tutor rekons its just - infinite
    i've tried dividing all by x cubed, factorizing and trying to cancel, multiplying but the conjugant and messing around with it

    here it is
    (lim as x approaches -3 from the right)

    lim 5x^2/ x^2 + 2x - 3
    x->-3+

    Thanks in advance
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  2. #2
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    Quote Originally Posted by Daniiel View Post
    Hey
    i've just been trying to work out this limit
    its been hours and my tutor rekons its just - infinite
    i've tried dividing all by x cubed, factorizing and trying to cancel, multiplying but the conjugant and messing around with it

    here it is
    (lim as x approaches -3 from the right)

    lim 5x^2/ x^2 + 2x - 3
    x->-3+

    Thanks in advance
    Dear Daniiel,

    By looking at the graph of this function you will understand it more clearly.
    Attached Thumbnails Attached Thumbnails annoying limit-sp1.png  
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  3. #3
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    Quote Originally Posted by Daniiel View Post
    Hey
    i've just been trying to work out this limit
    its been hours and my tutor rekons its just - infinite
    i've tried dividing all by x cubed, factorizing and trying to cancel, multiplying but the conjugant and messing around with it

    here it is
    (lim as x approaches -3 from the right)

    lim 5x^2/ x^2 + 2x - 3
    x->-3+

    Thanks in advance
    Try using long division:

    \frac{5x^2}{x^2 + 2x - 3} = 5 + \frac{15 - 10x}{x^2 + 2x - 3}.

    Now use partial fractions on the second term:

    \frac{A}{x + 3} + \frac{B}{x - 1} = \frac{15 - 10x}{x^2 + 2x - 3}

    \frac{A(x - 1) + B(x + 3)}{(x + 3)(x - 1)} = \frac{15 - 10x}{x^2 + 2x - 3}.

    Since the denominators are equal, so must be the numerators:

    A(x - 1) + B(x + 3) = 15 - 10x

    Ax - A + Bx + 3B = 15 - 10x

    3B - A + (A + B)x = 15 - 10x

    So 3B - A = 15 and A + B = -10

    Solving these equations simultaneously gives

    A = -\frac{45}{4} and B = \frac{5}{4}.


    Therefore

    5 + \frac{15 - 10x}{x^2 + 2x - 3} = 5 - \frac{45}{4(x + 3)} + \frac{5}{4(x - 1)}.


    Does this help?
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  4. #4
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    I did somthing similar to that but when you do lim x>-3 of the product you still get an undefined faction
    but i might be wrong

    does the graph show that as it approaches -3 that it isnt infinte?
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  5. #5
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    The important term is the -\frac{45}{4(x + 3)} = -\frac{45}{4}\left(\frac{1}{x + 3}\right).

    This is a basic hyperbola. You should know enough about hyperbolae of the form \frac{1}{x - a} to be able to say what happens as you approach a from each side.
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  6. #6
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    What you should have seen immediately is that the numerator, 5x^2 does not go to 0 at x= -3. Since the denominator does, that alone is enough to tell you that the limit does not exist and all your " dividing all by x cubed, factorizing and trying to cancel, multiplying by the conjugant and messing around with it" is useless. Saying it goes to -\infty is just saying it does not exist for a particular reason- and you can see that by looking at the graph or by evaluating the function at, say, -2.9.
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