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Math Help - trig limits...help?

  1. #1
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    trig limits...help?

    lim sin(2x)/tanx , x-->0

    I'm not sure if I'm doing this right, I've never used one of these before...hopefully I'm posting this in the right spot.
    Anyways, I can't seem to figure this question out...any help would be much appreciated!! Thanks
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  2. #2
    Eater of Worlds
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    \lim_{x\to 0}\frac{sin(2x)}{tan(x)}

    \lim_{x\to 0}\frac{2sin(x)cos(x)}{\frac{sin(x)}{cos(x)}}

    2\lim_{x\to 0}cos^{2}(x)=2
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  3. #3
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    Thanks for the quick reply...I'm just not quite sure how you got from one step to the next. I'm a little confused..?
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  4. #4
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    Two identities are used:
    • \sin (2\theta) = 2\sin \theta \cos \theta
    • \tan \theta = \frac{\sin \theta}{\cos \theta}


    So:
    \lim_{x \to 0} \frac{\sin 2x}{\tan x} = \lim_{x \to 0} \frac{\ 2\sin x \cos x \ }{\displaystyle \frac{\sin x}{\cos x}}

    Then do some algebra (dividing fractions is the same thing as multiplying the numerator by the reciprocal of the denominator):

    = \lim_{x \to 0 } 2\sin x \cos x \cdot \frac{\cos x}{\sin x}

    Things cancel and you should be good.
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  5. #5
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    where does the cos(x) come from in the numerator?
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  6. #6
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    Really not possible to explain it better...
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  7. #7
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    so it's impossible to explain how that cos(x) just popped up into the numerator?
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  8. #8
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    Not if you know that to divide by a fraction, you "invert and multiply"!

    \frac{a}{\frac{b}{c}}= a\left(\frac{c}{b}\right)= \frac{ac}{b}
    Do you understand that?

    When James Bond said he couldn't explain it better, he had already said that:
    "Then do some algebra (dividing fractions is the same thing as multiplying the numerator by the reciprocal of the denominator):"
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  9. #9
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    Well, thank you. That's all I needed.
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