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Math Help - Help with limits of indeterminant form.

  1. #1
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    Help with limits of indeterminant form.

    Dear all,
    Please help me for the next limits:
    1. lim_{x\rightarrow 0} \frac{x \sin{(\sin{x})}-\sin^{3}x}{x^6}

    2. lim_{x\rightarrow 0} \frac{1-\frac{x}{\ln{(1+x)}}}{x}

    3. lim_{x\rightarrow 0} \frac{(5+x)^x -5^x}{x^2}
    Last edited by mr fantastic; November 8th 2011 at 02:58 AM. Reason: Re-titled.
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  2. #2
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    Re: Please help me for the next limits

    Quote Originally Posted by matboy View Post
    Dear all,
    Please help me for the next limits:
    1. lim_{x\rightarrow 0} \frac{x \sin{(\sin{x})}-\sin^{3}x}{x^6}

    2. lim_{x\rightarrow 0} \frac{1-\frac{x}{\ln{(1+x)}}}{x}

    3. lim_{x\rightarrow 0} \frac{(5+x)^x -5^x}{x^2}
    What rules are you allowed to use?

    They are all indeterminate forms of "0/0" and L'Hosiptials rule could be used.
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  3. #3
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    Re: Please help me for the next limits

    Quote Originally Posted by TheEmptySet View Post
    What rules are you allowed to use?

    They are all indeterminate forms of "0/0" and L'Hosiptials rule could be used.
    Very tricky limits since they have been posted in the PRE-calculus subforum and therefore it is assumed that calculus cannot be used ....
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    Lightbulb Re: Please help me for the next limits

    Q2 can be easily solved by using L Hopital's rule but since calculus isnt allowed, I have used the series expansion of ln(1+x).

    \lim_{x \to 0}\frac{1-\frac{x}{\ln(x+1)}}{x}

    =\lim_{x \to 0}\frac{\ln{(1+x)}-x}{x \ln{(x+1)}}

    =\lim_{x \to 0}\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)-x}{x (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}

    =\lim_{x \to 0}\frac{(-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}{x (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}

    =\lim_{x \to 0}\frac{x^2(-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+...)}{x^2 (1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)}

    =\lim_{x \to 0}\frac{(-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+...)}{ (1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)}

    =\frac{(-\frac{1}{2}+\frac{0}{3}-\frac{0}{4}+...)}{ (1-\frac{0}{2}+\frac{0}{3}-\frac{0}{4}+...)}

    =-\frac{1}{2}
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  5. #5
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    Re: Please help me for the next limits

    Quote Originally Posted by sbhatnagar View Post
    Q2 can be easily solved by using L Hopital's rule but since calculus isnt allowed, I have used the series expansion of ln(1+x).

    \lim_{x \to 0}\frac{1-\frac{x}{\ln(x+1)}}{x}

    =\lim_{x \to 0}\frac{\ln{(1+x)}-x}{x \ln{(x+1)}}

    =\lim_{x \to 0}\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)-x}{x (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}

    =\lim_{x \to 0}\frac{(-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}{x (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}

    =\lim_{x \to 0}\frac{x^2(-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+...)}{x^2 (1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)}

    =\lim_{x \to 0}\frac{(-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+...)}{ (1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)}

    =\frac{(-\frac{1}{2}+\frac{0}{3}-\frac{0}{4}+...)}{ (1-\frac{0}{2}+\frac{0}{3}-\frac{0}{4}+...)}

    =-\frac{1}{2}
    It requires using Calculus to get the series expansion...
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  6. #6
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    Re: Please help me for the next limits

    Oh! sorry I completely forgot about that.
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    Re: Please help me for the next limits

    Quote Originally Posted by mr fantastic View Post
    Very tricky limits since they have been posted in the PRE-calculus subforum and therefore it is assumed that calculus cannot be used ....
    The OP didn't use an appropriate title or show any work. For that matter, he hasn't stopped by to "thank" anyone or clarify.

    So why are we assuming that he posted in the correct subforum?
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    Re: Please help me for the next limits

    Quote Originally Posted by TheChaz View Post
    The OP didn't use an appropriate title or show any work. For that matter, he hasn't stopped by to "thank" anyone or clarify.

    So why are we assuming that he posted in the correct subforum?
    I've assumed nothing. I'm making a point ......
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    Re: Please help me for the next limits

    Quote Originally Posted by mr fantastic View Post
    I've assumed nothing. I'm making a point ......
    i think we should take mr. f's suggestion and run with it. use only algebraic methods, and certain "primitive" limits like sin(x)/x as x --> 0 (which can be proved geometrically).

    yes, this may not be the original poster's intent, but are we mathematicians, or mice?
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