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Math Help - help me find this limit

  1. #1
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    help me find this limit

    im pretty sure im getting the right answer that there is no limit

    but the online math problem for my math class says i am wrong

    \frac{1}{t\sqrt{1+t}}-\frac{1}{t}



    thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by silencecloak View Post
    im pretty sure im getting the right answer that there is no limit

    but the online math problem for my math class says i am wrong

    \frac{1}{t\sqrt{1+t}}-\frac{1}{t}



    thanks
    What is the value that t is approaching?

    --Chris
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    What is the value that t is approaching?

    --Chris
    Well I thought, nothing

    But the site says I'm wrong


    I don't think I answered your question, but I don't know how to
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  4. #4
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    Quote Originally Posted by silencecloak View Post
    Well I thought, nothing

    But the site says I'm wrong


    I don't think I answered your question, but I don't know how to
    If you're being asked to find a limit then you surely have been told the value that t is approaching?? Is this what your're trying to find:

    \lim_{t \rightarrow 0} \left( \frac{1}{t\sqrt{1+t}}-\frac{1}{t}\right).

    The value of the limit is - \frac{1}{2}.


    Note that \frac{1}{t\sqrt{1+t}}-\frac{1}{t} = \frac{1 - \sqrt{1+t}}{t \sqrt{1 + t}}.

    Multiplying this expression by 1 = \frac{1 + \sqrt{1 + t} }{1 + \sqrt{1 + t}}, simplify the numerator, cancel the common factor of t and then try taking the limit again.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    If you're being asked to find a limit then you surely have been told the value that t is approaching?? Is this what your're trying to find:

    \lim_{t \rightarrow 0} \left( \frac{1}{t\sqrt{1+t}}-\frac{1}{t}\right).

    The value of the limit is - \frac{1}{2}.


    Note that \frac{1}{t\sqrt{1+t}}-\frac{1}{t} = \frac{1 - \sqrt{1+t}}{t \sqrt{1 + t}}.

    Multiplying this expression by 1 = \frac{1 + \sqrt{1 + t} }{1 + \sqrt{1 + t}}, simplify the numerator, cancel the common factor of t and then try taking the limit again.

    oh, im sorry Chris i misunderstood your question

    and I still don't see where you are getting 1/2 from

    I always end up with t(t+1) in the denominator which when u plug in 0....is 0

    \frac{1}{t\sqrt{1+t}}-\frac{1}{t} = \frac{x - \sqrt{1+t}}{(x)t \sqrt{1 + t}}

    you can cancel those two "x" out? seems weird to me
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  6. #6
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    Quote Originally Posted by silencecloak View Post
    oh, im sorry Chris i misunderstood your question

    and I still don't see where you are getting 1/2 from

    I always end up with t(t+1) in the denominator which when u plug in 0....is 0

    \frac{1}{t\sqrt{1+t}}-\frac{1}{t} = \frac{x - \sqrt{1+t}}{(x)t \sqrt{1 + t}}

    you can cancel those two "x" out? seems weird to me
    \frac{1 - \sqrt{1+t}}{t \sqrt{1 + t}}


    = \frac{(1 - \sqrt{1+t})}{t \sqrt{1 + t}} \times \frac{(1 + \sqrt{1 + t}) }{(1 + \sqrt{1 + t)}}


    = \frac{(1 - \sqrt{1+t}) (1 + \sqrt{1 + t}) }{t \sqrt{1 + t} (1 + \sqrt{1 + t}) }

    =  \frac{-t}{t \sqrt{1 + t} (1 + \sqrt{1 + t}) }.


    Quote Originally Posted by mr fantastic View Post
    [snip]
    Multiplying this expression by 1 = \frac{1 + \sqrt{1 + t} }{1 + \sqrt{1 + t}}, Mr F says: Done.

    simplify the numerator, Mr F says: Done.

    cancel the common factor of t and then try taking the limit again. Mr F says: Left for you to do.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    \frac{1 - \sqrt{1+t}}{t \sqrt{1 + t}}


    = \frac{(1 - \sqrt{1+t})}{t \sqrt{1 + t}} \times \frac{(1 + \sqrt{1 + t}) }{(1 + \sqrt{1 + t)}}


    = \frac{(1 - \sqrt{1+t}) (1 + \sqrt{1 + t}) }{t \sqrt{1 + t} (1 + \sqrt{1 + t}) }

    =  \frac{-t}{t \sqrt{1 + t} (1 + \sqrt{1 + t}) }.
    Thank you for helping me
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