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Math Help - Limit as x->0 of a complicated trig function

  1. #1
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    Limit as x->0 of a complicated trig function

    The function is:

    (tan(x))^3 - x
    x+x^2
    Allowed facts to use are that the limit as x tends to 0 of sin(x)/x=1, and theorems including the algebra of limits, sandwich theorem and the 'changing the variables' theorem.



    I believe the limit to be -1, and I'm confident of showing it using the thorems stated above, so my problem isn't specifically with showing the limit, it's more with formatting the problem. Re-writing the initial function in such a way that I can use the limit of sin(x)/x.

    I started by taking (tanx)^3= (sin(x)^3)/(cos(x)^3), and then (cos(x))^3= cos(x)(1-(sin(x))^2). But to be honest it all gets rather complex, and as I haven't learnt Latex yet it would look an absolute mess on here, and also my trigonometry is a bit rusty, so if anyone could help me through this initial part, or offer an easier method, I'd be grateful. Thanks!
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  2. #2
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    Re: Limit as x->0 of a complicated trig function

    Quote Originally Posted by struggler View Post
    The function is:

    (tan(x))^3 - x
    x+x^2
    Allowed facts to use are that the limit as x tends to 0 of sin(x)/x=1, and theorems including the algebra of limits, sandwich theorem and the 'changing the variables' theorem.



    I believe the limit to be -1, and I'm confident of showing it using the theorems stated above, so my problem isn't specifically with showing the limit, it's more with formatting the problem. Re-writing the initial function in such a way that I can use the limit of sin(x)/x.

    I started by taking (tanx)^3= (sin(x)^3)/(cos(x)^3), and then (cos(x))^3= cos(x)(1-(sin(x))^2). But to be honest it all gets rather complex, and as I haven't learned Latex yet it would look an absolute mess on here, and also my trigonometry is a bit rusty, so if anyone could help me through this initial part, or offer an easier method, I'd be grateful. Thanks!
    Yes, the limit is -1.

    Probably the easiest way to show this is with Taylor series. I suppose the is not allowed for you.

    After taking (tanx)^3= (\sin(x)^3)/(\cos(x)^3), multiply the whole expression by \frac{\displaystyle \frac{\cos(x)}{x}}{\displaystyle \frac{\cos(x)}{x}}\ .
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  3. #3
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    Re: Limit as x->0 of a complicated trig function

    Equivalently, write tan^3(x)= \frac{sin^3(x)}{cos^3(x)} and x^2+1= x(x+1). Now you can write the fraction as
    \frac{sin(x)}{x}\frac{sin^2(x)}{cos^3(x)(x+1)}- \frac{1}{x^2+1}
    and take the limit of each fraction separately.
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  4. #4
    Grand Panjandrum
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    Re: Limit as x->0 of a complicated trig function

    You need to know that:

    \lim_{x \to 0} \frac{\tan(x)}{x}=1

    which you can derive from the equivalent limit with \sin(x) in place of \tan(x).

    Then:

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

    CB (posting a full solution because after 10 days off line this can no longer be handed in for credit)
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  5. #5
    Member sbhatnagar's Avatar
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    Lightbulb Re: Limit as x->0 of a complicated trig function

    There is another way to do it.

    It is known that that as x\to 0, \sin{x}\approx x and \cos{x}\approx 1-\frac{x^2}{2}.

    Limit becomes: \\ \lim_{x\to 0}\frac{\tan^3{x}-x}{x+x^2}=\lim_{x\to 0}\frac{\frac{\sin^3x}{\cos^3x}-x}{x+x^2}=\lim_{x\to 0}\frac{\frac{8x^3}{(2-x^2)^3}-x}{x+x^2}=\lim_{x\to 0}\frac{\frac{8x^2}{(2-x^2)^3}-1}{x+1}=-1
    Last edited by sbhatnagar; November 5th 2011 at 04:23 AM.
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  6. #6
    Grand Panjandrum
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    Re: Limit as x->0 of a complicated trig function

    Quote Originally Posted by sbhatnagar View Post
    There is another way to do it.

    It is known that that as x\to 0, \sin{x}\approx x and \cos{x}\approx 1-\frac{x^2}{2}.

    Limit becomes: \\ \lim_{x\to 0}\frac{\tan^3{x}-x}{x+x^2}=\lim_{x\to 0}\frac{\frac{\sin^3x}{\cos^3x}-x}{x+x^2}=\lim_{x\to 0}\frac{\frac{8x^3}{(2-x^2)^3}-x}{x+x^2}=\lim_{x\to 0}\frac{\frac{8x^2}{(2-x^2)^3}-1}{x+1}=-1
    Without making precise what you mean by " \approx" this does not work as it is. My post uses this idea, but in the form of the limit of tan(x) over x, which does work without further ado.

    CB
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  7. #7
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    Re: Limit as x->0 of a complicated trig function

    Thanks for all your advice everyone, I used something along these lines in the end
    Quote Originally Posted by HallsofIvy View Post
    Equivalently, write tan^3(x)= \frac{sin^3(x)}{cos^3(x)} and x^2+1= x(x+1). Now you can write the fraction as
    \frac{sin(x)}{x}\frac{sin^2(x)}{cos^3(x)(x+1)}- \frac{1}{x^2+1}
    and take the limit of each fraction separately.
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