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Math Help - square root limit

  1. #1
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    square root limit

    lim_{x->0}\frac{\sqrt{x^2+arctanx}}{x}

    when we put the x in the denominator inside the root
    we get absolute value so wee need to split into cases
    let us deal with only one case

    lim_{x->0^+}\frac{\sqrt{1+\frac{arctan}{x^2}}}{1}
    so i have a member in side the root which is 0/0

    can i solve this thing without lhopital?
    Last edited by transgalactic; September 17th 2011 at 09:58 PM.
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  2. #2
    Grand Panjandrum
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    Re: square root limit

    Quote Originally Posted by transgalactic View Post
    lim_{x->0}\frac{\sqrt{x^2+arctanx}}{x}

    when we put the x in the denominator inside the root
    we get absolute value so wee need to split into cases
    let us deal with only one case

    lim_{x->0^+}\frac{\sqrt{1+\frac{arctan}{x^2}}}{1}
    so i have a member in side the root which is 0/0

    can i solve this thing without lhopital?
    \lim_{x\to 0} \frac{\arctan(x)}{x}=1

    so:

    \lim_{x\to 0} \frac{\arctan(x)}{x^2}=\infty

    CB
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  3. #3
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    Re: square root limit

    Quote Originally Posted by CaptainBlack View Post
    \lim_{x\to 0} \frac{\arctan(x)}{x}=1

    so:

    \lim_{x\to 0} \frac{\arctan(x)}{x^2}=\infty

    CB
    why
    \lim_{x\to 0} \frac{\arctan(x)}{x}=1
    ??
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  4. #4
    Grand Panjandrum
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    Re: square root limit

    Quote Originally Posted by transgalactic View Post
    why
    \lim_{x\to 0} \frac{\arctan(x)}{x}=1
    ??
    For small x we have \tan(x)\sim x, so put y=\tan(x) and as x is small so is y and vice-versa and so for small y:

    y\sim \arctan(y)

    CB
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  5. #5
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    Re: square root limit

    its like sinx/x

    its a definition
    ?
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  6. #6
    Grand Panjandrum
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    Re: square root limit

    Quote Originally Posted by transgalactic View Post
    its like sinx/x

    its a definition
    ?
    It is not a definition, it is a consequence of the definition of the tan/arctan function/s.

    CB
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  7. #7
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    Re: square root limit

    so i cant use it if its not a definition
    and i cant use your explanation because its based on approximation.

    and i cant do lopital only on one fracture

    so how to solve it
    ?
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  8. #8
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    Re: square root limit

    Why can't you use it if it is not a definition? tan(x)= sin(x)/cos(x) so tan(x)/x= (sin(x)/x) cos(x). Now, both of those limits, as x goes to 0, should be easy.
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  9. #9
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    Re: square root limit

    but its arctanx not tan

    ?
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