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Math Help - Limit

  1. #1
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    Limit

    OK I am getting a different answer then the book.
    Find the limit
    x------>4
    square root(x-3) - 1 / x-4
    X-3 is the only part in the fraction that is square rooted.
    I know you can bring it down and be lefted with
    1/square (x-3) which gives you lim X=1 back the books says its one half so im doing something wrong .
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Nightasylum View Post
    OK I am getting a different answer then the book.
    Find the limit
    x------>4
    square root(x-3) - 1 / x-4
    X-3 is the only part in the fraction that is square rooted.
    I know you can bring it down and be lefted with
    1/square (x-3) which gives you lim X=1 back the books says its one half so im doing something wrong .
    \lim_{x\to4}\frac{\sqrt{x-3}-1}{x-4}=\lim_{x\to4}\frac{\sqrt{x-3}-1}{x-4}\cdot\frac{\sqrt{x-3}+1}{\sqrt{x-3}+1} =\lim_{x\to4}\frac{x-3-1}{\left(x-4\right)\left(\sqrt{x-3}+1\right)}=\lim_{x\to4}\frac{1}{\sqrt{x-3}+1}

    I'm sure you can take it from here...
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  3. #3
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    Ok so you acually rationalize the whole numerator? I was just doing the square root part.
    Also can you help with
    X---->-5
    x^3+125/x+5
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Nightasylum View Post
    Ok so you acually rationalize the whole numerator? I was just doing the square root part.
    Also can you help with
    X---->-5
    x^3+125/x+5
    You will need to use the formula for factoring the sum of two cubes:

    a^3+b^3=(a+b)(a^2-ab+b^2)
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  5. #5
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    (x+5)(x^2-5x+5^2)/(x+5) = x^2-5x+5^2 = 75
    I think I did that right I need to remember that formula thanks alot.
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Nightasylum View Post
    (x+5)(x^2-5x+5^2)/(x+5) = x^2-5x+5^2 = 75
    I think I did that right I need to remember that formula thanks alot.


    \lim_{x\to-5}\frac{x^3+125}{x+5}=75
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