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Math Help - Evaluating Limits

  1. #1
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    Evaluating Limits

    I need to evaluate the following limits:
    1) \lim_{x\to0} (\sqrt{x+1})/x, where (x>-1)
    2) \lim_{x\to\infty} (\sqrt{x+1})/x, where (x>0)

    here is what I have so far
    (\sqrt{x+1})/x = \sqrt{\frac{x+1}{x^2}} = \sqrt{\frac{1}{x} + \frac{1}{x^2}} < \sqrt{\frac{2}{x}} . . . I don't know if I'm on the right track or not. Any help is greatly appreciated,thanks
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  2. #2
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    Quote Originally Posted by CrazyCat87 View Post
    I need to evaluate the following limits:
    1) \lim_{x\to0} (\sqrt{x+1})/x, where (x>-1)
    2) \lim_{x\to\infty} (\sqrt{x+1})/x, where (x>0)

    here is what I have so far
    (\sqrt{x+1})/x = \sqrt{\frac{x+1}{x^2}} = \sqrt{\frac{1}{x} + \frac{1}{x^2}} < \sqrt{\frac{2}{x}} . . . I don't know if I'm on the right track or not. Any help is greatly appreciated,thanks
    = \sqrt{\frac{1}{x} + \frac{1}{x^2}}

    Using that you can conclude. If x is infinity, your fractions will be 0. If x is 0, your fractions will be infinitly big, infinity.
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  3. #3
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    Quote Originally Posted by felper View Post
    = \sqrt{\frac{1}{x} + \frac{1}{x^2}}

    Using that you can conclude. If x is infinity, your fractions will be 0. If x is 0, your fractions will be infinitly big, infinity.
    Yea I undertsand that but I am required to prove these statements strictly by definition. Otherwise it'd be easy!
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  4. #4
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    Then you should have said that! What you said was "I need to evaluate the following limits".
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Then you should have said that! What you said was "I need to evaluate the following limits".
    Yea that's my fault, our ta says that but expects us to have a full proof lol
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