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Math Help - limits! : )

  1. #1
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    limits! : )

    Find the following limits:

    <br /> <br />
\lim_{x \to 0} \frac{tan 2x}{3x cosx}<br />

    <br /> <br />
\lim_{x \to 0} x^3 sin \frac{2}{x}<br />

    mhmmmm help kthx : )
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  2. #2
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    1. \tan 2x = \frac{2 \tan x}{1 - \tan^2 x}

    \frac{\tan 2x}{3x \cos x} =

    \frac{2 \tan x}{3x \cos x (1 - \tan^2 x)} =

    \frac{2 \sin x}{3x \cos^2 x (1 - \tan^2 x)} =

    \frac{2}{3} \cdot \frac{\sin x}{x} \cdot \frac{1}{\cos^2 x (1 - \tan^2 x)}

    The left limit is 2/3, the middle limit is 1, and the right limit is 1. Therefore, the limit of the product is 2/3.

    2. x^3 \sin \left(\frac{2}{x}\right)

    The limit of x^3 is 0, and the sine function is bounded between -1 and 1, so the limit is 0.
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  3. #3
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    ok so like

    <br /> <br />
\lim_{x \to 0^-} \frac{5}{csc x}<br />

    that equals 0?
    because..

    <br /> <br />
\frac{5}{csc x} = 5 \cdot sinx = 5 sin(0) = 5 \cdot 0 = 0<br />
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  4. #4
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    Quote Originally Posted by iz1hp View Post
    ok so like

    <br /> <br />
\lim_{x \to 0^-} \frac{5}{csc x}<br />

    that equals 0?
    because..

    <br /> <br />
\frac{5}{csc x} = 5 \cdot sinx = 5 sin(0) = 5 \cdot 0 = 0<br />
    Correct.
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  5. #5
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    excellent : ) im grasping the concept hahaha thanks!
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