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Math Help - Finding limits

  1. #1
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    Exclamation Finding limits

    If I want to find the \lim_{x\rightarrow 0}\frac {xcosx -sin(x)}{x^2}, using L'Hopital's rule, also knowing \lim_{x\rightarrow 0}\frac{sinx}{x} is equal to 1, how many times should I differentiate?

    i differentiated it once, and my answer was that the limit = -1

    Is this correct?

    Thanks in advance.
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  2. #2
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    I do not find the same result as you, apart from the fact that I needed only to differentiate once !
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  3. #3
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    \lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2} differentiate once:

    \lim_{x\rightarrow 0}\frac{cosx-xsinx-cosx}{2x} so

    \lim_{x\rightarrow 0}\frac{-xsinx}{2x} eliminate x (which you didnīt) and you have that the limit is 0
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  4. #4
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    Quote Originally Posted by Inti View Post
    \lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2} differentiate once:

    \lim_{x\rightarrow 0}\frac{cosx-xsinx-cosx}{2x} so

    \lim_{x\rightarrow 0}\frac{-xsinx}{2x} eliminate x (which you didnīt) and you have that the limit is 0
    Hey, thanks for the quick reply.

    I can't see how \frac{-xsinx}{2x} is the derivative of [tex]\frac{cosx-xsinx-cosx}{2x}[/math. Could you please explain?

    Thanks again.
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  5. #5
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    I just cancelled the opposite cosines, I wasnīt differentiating...
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  6. #6
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    Quote Originally Posted by Inti View Post
    I just cancelled the opposite cosines, I wasnīt differentiating...
    Sorry Inti, but I still can't see what you're doing...

    Initially we had:

    \lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2}

    Then you differentiate it, right? or what did you do to get the following?

    \lim_{x\rightarrow 0}\frac{cosx-xsinx-cosx}{2x}

    I can see how you can cancel cosx after differentiating...

    See, the problem is now that my derivative of \lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2} is \lim_{x\rightarrow 0}\frac{-x^2sinx+2xcosx-2sinx}{x^3}
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  7. #7
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    Quote Originally Posted by tsal15 View Post
    Sorry Inti, but I still can't see what you're doing...

    Initially we had:

    \lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2}

    Then you differentiate it, right? or what did you do to get the following?

    \lim_{x\rightarrow 0}\frac{cosx-xsinx-cosx}{2x}

    I can see how you can cancel cosx after differentiating...

    See, the problem is now that my derivative of \lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2} is \lim_{x\rightarrow 0}\frac{-x^2sinx+2xcosx-2sinx}{x^3}
    OK tsal15 just a minute
    Differentiate \frac{xcosx-sinx}{x^2} is \frac{-x^2sinx-2xcosx+2sinx}{x^3} (sign mistake I think)

    Anyway it does not matter because this is not what L'Hopital's rule says
    It says
    \lim_{x\rightarrow 0}\frac{f(x)}{g(x)}\:=\:\lim_{x\rightarrow 0}\frac{f'(x)}{g'(x)}

    You do not need to differentiate \frac{f(x)}{g(x)}

    Therefore
    \lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2} is \lim_{x\rightarrow 0}\frac{-xsinx}{2x}
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  8. #8
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    **** yeah!!! I was going to ask exactly that: if he/she was differentiating THE ENTIRE FUNCTION instead of the dividend and the divisor separately...
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  9. #9
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    Quote Originally Posted by Inti View Post
    **** yeah!!! I was going to ask exactly that: if he/she was differentiating THE ENTIRE FUNCTION instead of the dividend and the divisor separately...
    I get you now...So, the limit is equal to ZERO?

    thanks for your help, Inti
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  10. #10
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    Quote Originally Posted by running-gag View Post
    OK tsal15 just a minute
    Differentiate \frac{xcosx-sinx}{x^2} is \frac{-x^2sinx-2xcosx+2sinx}{x^3} (sign mistake I think)

    Anyway it does not matter because this is not what L'Hopital's rule says
    It says
    \lim_{x\rightarrow 0}\frac{f(x)}{g(x)}\:=\:\lim_{x\rightarrow 0}\frac{f'(x)}{g'(x)}

    You do not need to differentiate \frac{f(x)}{g(x)}

    Therefore
    \lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2} is \lim_{x\rightarrow 0}\frac{-xsinx}{2x}
    OF COURSE!!! DOH!!! thanks, running-gag. so, the limit is equal to 0 (zero)?
    thanks for your help.
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  11. #11
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    Quote Originally Posted by tsal15 View Post
    I get you now...So, the limit is equal to ZERO?
    Yes the limit is equal to 0
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  12. #12
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    Quote Originally Posted by running-gag View Post
    Yes the limit is equal to 0
    GREAT thanks
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