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Math Help - Limit headaches

  1. #1
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    Limit headaches

    Ok, so here is the limit question that has been giving me some difficulty:

    The lim of x+2 divided by the square root of 6 + x minus the square root of 2 - x as x approaches -2

    (Sorry, if I don't know how to put it in symbols otherwise I would >//<; )

    Also, any general tips on finding the proper way to "play with"/solve limits would be much appreciated. Thanks for any help in advance!
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  2. #2
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    Quote Originally Posted by Toaster View Post
    Ok, so here is the limit question that has been giving me some difficulty:

    The lim of x+2 divided by the square root of 6 + x minus the square root of 2 - x as x approaches -2

    (Sorry, if I don't know how to put it in symbols otherwise I would >//<; )

    Also, any general tips on finding the proper way to "play with"/solve limits would be much appreciated. Thanks for any help in advance!
    \lim_{x \rightarrow -2} \frac{x+2}{\sqrt{6+x} - \sqrt{2-x}}.

    Multiply the expression by 1 = \frac{ \sqrt{6+x} + \sqrt{2-x} }{\sqrt{6+x} + \sqrt{2-x}} and simplify the denominator. You should get:

    \lim_{x \rightarrow -2} \frac{(x+2)(\sqrt{6+x} + \sqrt{2-x})}{4 + 2x}.

    Cancel the obvious common factor and try taking the limit again.
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  3. #3
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    Ah! Thank you! I see I shouldn't have multiplied the top out, do you think you could help me with another question?

    The limit of x-8 divided by (the cube root of x) - 2 as x approaches 8.
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  4. #4
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    Quote Originally Posted by Toaster View Post
    Ah! Thank you! I see I shouldn't have multiplied the top out, do you think you could help me with another question?

    The limit of x-8 divided by (the cube root of x) - 2 as x approaches 8.
    Hint: x - 8 = (x^{1/3})^3 - 2^3. You know how to factorise a difference of two cubes, right ....?


    FYI New threads should be created for new questions.
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  5. #5
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    Mmm, yes I do.

    Thanks for the hint and the heads-up. ^_^
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