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Math Help - Trig. Limits

  1. #1
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    Trig. Limits

    a) Calculate the limit value to \frac{sinx-x}{x(cosx-1)} when x\rightarrow0

    The point is to use Taylor or McLaurin polynomial..?.

    B) Make a smilar task that will equal \frac{21}{5}
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  2. #2
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    Quote Originally Posted by lynch-mob View Post
    a) Calculate the limit value to \frac{sinx-x}{x(cosx-1)} when x\rightarrow0

    The point is to use Taylor or McLaurin polynomial..?.

    B) Make a smilar task that will equal \frac{21}{5}
    Have you tried substituting the Maclaurin Series for sin x and cos x and then simplifying the resulting expression?
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  3. #3
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    too hard

    Quote Originally Posted by mr fantastic View Post
    Have you tried substituting the Maclaurin Series for sin x and cos x and then simplifying the resulting expression?
    The Mclaurin series for sinx is something like \frac{x^{2n-1}}{(2n-1)!}

    and for cos something like \frac{x^{2n}}{2n!}

    So should I put these straight into the formula replacing sin and cos?
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  4. #4
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    Quote Originally Posted by lynch-mob View Post
    The Mclaurin series for sinx is something like \frac{x^{2n-1}}{(2n-1)!}

    and for cos something like \frac{x^{2n}}{2n!}

    So should I put these straight into the formula replacing sin and cos?
    Yes, but only the first couple of terms of each are necessary.
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  5. #5
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    Quote Originally Posted by lynch-mob View Post
    a) Calculate the limit value to \frac{sinx-x}{x(cosx-1)} when x\rightarrow0

    The point is to use Taylor or McLaurin polynomial..?.

    B) Make a smilar task that will equal \frac{21}{5}
    \sin(x)-x=\bigg[x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\bigg]=\frac{-x^3}{3!}+\frac{x^5}{5!}-...\sim\frac{-x^3}{3!}

    and x(\cos(x)-1)=x\bigg[1-\frac{x^2}{2!}+\frac{x^4}{4!}-...-1\bigg]=x\bigg[\frac{-x^2}{2!}+\frac{x^4}{4!}-...\bigg]=\frac{-x^3}{2!}+\frac{x^5}{4!}-...\sim\frac{-x^3}{2!}


    So we have that \lim_{x\to{0}}\frac{\sin(x)-x}{x(\cos(x)-1)}\sim\lim_{x\to{0}}\frac{\frac{-x^3}{3!}}{\frac{-x^2}{2!}}=\frac{1}{3}
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    \sin(x)-x=\bigg[x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\bigg]=\frac{-x^3}{3!}+\frac{x^5}{5!}-...\sim\frac{-x^3}{3!}

    and x(\cos(x)-1)=x\bigg[1-\frac{x^2}{2!}+\frac{x^4}{4!}-...-1\bigg]=x\bigg[\frac{-x^2}{2!}+\frac{x^4}{4!}-...\bigg]=\frac{-x^3}{2!}+\frac{x^5}{4!}-...\sim\frac{-x^3}{2!}


    So we have that \lim_{x\to{0}}\frac{\sin(x)-x}{x(\cos(x)-1)}\sim\lim_{x\to{0}}\frac{\frac{-x^3}{3!}}{\frac{-x^2}{2!}}=\frac{1}{3}
    Im sorry but I donīt understand much of this? Isnīt the point when calculating with lim that if x\rightarrow0 then you put x=0 in the expression. When i do that i get only zeros.

    And what happens to the x and 1 in x(cosx-1)? Same with the -x in sinx - x??

    And how do you get that the cosx (mclaurin...) expression is approximately \frac{-x^3}{2!} and the same with the sin expression?

    Hope you have time for at least a brief explanation.

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  7. #7
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    Quote Originally Posted by lynch-mob View Post
    Im sorry but I donīt understand much of this? Isnīt the point when calculating with lim that if x\rightarrow0 then you put x=0 in the expression. When i do that i get only zeros.

    And what happens to the x and 1 in x(cosx-1)? Same with the -x in sinx - x??

    And how do you get that the cosx (mclaurin...) expression is approximately \frac{-x^3}{2!} and the same with the sin expression?

    Hope you have time for at least a brief explanation.

    I have done some editing to Mathstud28's reply, but you should have realised (and in fact done) these things yourself:

    Quote Originally Posted by Mathstud28 and slightly edited by Mr F to save reinventing the wheel View Post

    \sin(x)-x=\bigg[x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\bigg]{\color{red} - x} =\frac{-x^3}{3!}+\frac{x^5}{5!}-... Mr F says: The bit in red was missing. Do you understand this part?

    and x(\cos(x)-1)=x\bigg[1-\frac{x^2}{2!}+\frac{x^4}{4!}-...-1\bigg]=x\bigg[\frac{-x^2}{2!}+\frac{x^4}{4!}-...\bigg]=\frac{-x^3}{2!}+\frac{x^5}{4!} -... Mr F says: Do you understand this part?


    So we have that \lim_{x\to{0}}\frac{\sin(x)-x}{x(\cos(x)-1)} = \lim_{x\to{0}}\frac{ \frac{-x^3}{3!} + \frac{x^5}{5!} -...}{ \frac{-x^{\color{red}3}}{2!} + \frac{x^5}{5!} -...} Mr F says: The bit in red is a correction.
    Now divide numerator and denominator by x^3:

    = \lim_{x\to{0}}\frac{ \frac{-1}{3!} + \frac{x^2}{5!} -...}{ \frac{-1}{2!} + \frac{x^2}{5!} -...}

    Now "put x = 0 in the expression" as you say it: = \frac{\frac{-1}{3!}}{\frac{-1}{2!}} = \frac{2!}{3!}  = \frac{1}{3}
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  8. #8
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    Reinventing the wheel

    Ok now I get it. Sorry guys for being slow.

    I missed a few parts which totally jammed me.

    I canīt praise this site and these people helping here enough. Thanks!!
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