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Math Help - Factoring an X out

  1. #1
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    Factoring an X out

    I am stuck on the algebra in a basic limit problem example.

    (sqrt of x^2 + x) + x

    They factor out an x and get

    x * [(sqrt of 1 + 1/x) + 1]

    This is just the denominator of the problem but it is the part I dont understand.
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    \displaystyle \sqrt{x^2 + x} + x = \sqrt{x^2\left(1 + \frac{1}{x}\right)} + x

    \displaystyle = \sqrt{x^2}\sqrt{1 + \frac{1}{x}} + x

    \displaystyle = x\sqrt{1 + \frac{1}{x}} + x

    \displaystyle = x\left(\sqrt{1+\frac{1}{x}} + 1\right).
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    Thanks. Makes sense now that I see what you did. I dont think I would have ever thought to do the first step. Guess I need more practise.
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    When doing this:
    \displaystyle \sqrt{x^2 + x} + x = \sqrt{x^2\left(1 + \frac{1}{x}\right)} + x

    Why does the x turn into (1+\frac{1}{x})?

    I can't see it...
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  5. #5
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    What's \displaystyle x \div x^2?
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    \frac{1}{x}
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  7. #7
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    Quote Originally Posted by MaverickUK82 View Post
    Why does the x turn into (1+\frac{1}{x})?
    Well of course it does not do that.
    Do you see that \displaystyle x^2\left(1 + \frac{1}{x}\right)=x^2+x~?
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    Yes, I totally understand now.

    Had a blank moment there and didn't fully understand (or read) the question...
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    Quote Originally Posted by Prove It View Post
    \displaystyle \sqrt{x^2 + x} + x = \sqrt{x^2\left(1 + \frac{1}{x}\right)} + x

    \displaystyle = \sqrt{x^2}\sqrt{1 + \frac{1}{x}} + x

    \displaystyle = x\sqrt{1 + \frac{1}{x}} + x

    \displaystyle = x\left(\sqrt{1+\frac{1}{x}} + 1\right).
    It should be noted that \sqrt{x^2} = |x| not x. Since the OP said "I am stuck on the algebra in a basic limit problem example." this fact is likely to be important in getting the correct answer to the actual question.
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    Quote Originally Posted by mr fantastic View Post
    It should be noted that \sqrt{x^2} = |x| not x. Since the OP said "I am stuck on the algebra in a basic limit problem example." this fact is likely to be important in getting the correct answer to the actual question.
    Thats is a good point. I always space that fact. The original question in the book was a limit a infinitey with the problem I stated in the numerator. They use the conjugate and in the end cancel out the X top and bottom and end up with an answer of 1/2. If I new Latex it would be more apparent. Dont know if the limit at infinitey has anything to do with not needing the absolute value.
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  11. #11
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    Since you're going to a positive value (I assume you mean \displaystyle \lim_{x \to +\infty}), you should note that \displaystyle |x| = x, so the modulus signs would not be necessary in this case...
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  12. #12
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    Quote Originally Posted by Prove It View Post
    Since you're going to a positive value (I assume you mean \displaystyle \lim_{x \to +\infty}), you should note that \displaystyle |x| = x, so the modulus signs would not be necessary in this case...
    A limit question requiring this sort of algebraic re-arrangement only makes sense if x is approaching \displaystyle -\infty. It's trivial if \displaystyle x \to +\infty and no algebraic arrangement is necessary.

    Again we suffer from the typical ambiguity that arises when the real question does not get posted by the OP.
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  13. #13
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    Quote Originally Posted by mr fantastic View Post
    A limit question requiring this sort of algebraic re-arrangement only makes sense if x is approaching \displaystyle -\infty.
    What about \displaystyle \lim _{x \to \infty } \sqrt {x^2  + x}  - x~?
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    Quote Originally Posted by Plato View Post
    What about \displaystyle \lim _{x \to \infty } \sqrt {x^2 + x} - x~?
    That was the original question. I thought it might be more cumbersome to try and write the question without latex and the limit part seemed simple enough. I just got stumped on the algebra. Thanks for everyones help.
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  15. #15
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    Quote Originally Posted by mr fantastic View Post
    A limit question requiring this sort of algebraic re-arrangement only makes sense if x is approaching \displaystyle -\infty. It's trivial if \displaystyle x \to +\infty and no algebraic arrangement is necessary.

    Again we suffer from the typical ambiguity that arises when the real question does not get posted by the OP.
    My question was answered in the first reply and might I add very quickly which was greatly appreciated. I only put part of the question in because it was the only part I didnt understand and seemed to fit in the algebra section instead of the calculus section. Sorry for the confusion.
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