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Thread: Limit.

  1. #1
    Super Member Showcase_22's Avatar
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    Limit.

    Find the value of $\displaystyle \lim_{x \rightarrow 0} \frac{sin(x)-x}{tan(x)-x}$.
    This is what I did:

    $\displaystyle f(x)=sin(x)-x$
    $\displaystyle f'(x)=cos(x)-1$
    $\displaystyle f^{(2)}(x)=-sin(x)$
    $\displaystyle f^{(3)}(x)=-cos(x)$

    $\displaystyle g(x)=tan(x)-x$
    $\displaystyle g'(x)=sec^2(x)-1$
    $\displaystyle g^{(2)}(x)=\frac{2sin(x)}{cos^3(x)}$
    $\displaystyle g^{(3)}(x)=\frac{2cos^2(x)+6sin^2(x)}{cos^4(x)}$

    Hence:

    $\displaystyle a_3=f^{(3)}(0)=-cos(0)=-1$
    $\displaystyle b_3=g^{(3)}(0)=2$

    Hence $\displaystyle \lim_{x \rightarrow 0} \frac{sin(x)-x}{tan(x)-x}=\frac{a_3}{b_3}=-\frac{1}{2}$

    However, when I put x=0.0000000001 into my calculator (as a check) I end up with 1.

    What's going wrong??? :S
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  2. #2
    Super Member redsoxfan325's Avatar
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    Swampscott, MA
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    According to Maple 12, $\displaystyle -\frac{1}{2}$ is correct. Also, just looking at the graph of the function, it appears that $\displaystyle f(x) = \frac{sin(x)-x}{tan(x)-x}$ is continuous at $\displaystyle x = 0$. I think your mistake occurred when you plugged in 0.0000000001.
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  3. #3
    Super Member Showcase_22's Avatar
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    Oh right.

    It's a rather strange function since putting 0.01 into the function seems to suggest it tends to 0, but putting something much smaller in seems to suggest it tends to 1!

    Well I guess there's no substitute for analytical methods!

    thanks!
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