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Math Help - limits problems--need help!

  1. #1
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    limits problems--need help!

    I'm stuck on getting the limits for sin/cos problems... mind helping me solve these problems?

    1. lim x-->0 x^2/1-cosx
    2. lim x-->0 x/sin3x
    3. lim x-->0 sin5x/sin2x

    (I don't know how to type limit problems in this forum..)
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  2. #2
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    Are you allowed to use L'Hospital's Rule? If so,

    \lim_{x \to 0} \frac{x^2}{1 - \cos x} =

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

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

    You can use the same approach for the other two problems.
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  3. #3
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    Quote Originally Posted by kdl00 View Post
    I'm stuck on getting the limits for sin/cos problems... mind helping me solve these problems?

    1. lim x-->0 x^2/1-cosx
    2. lim x-->0 x/sin3x
    3. lim x-->0 sin5x/sin2x

    (I don't know how to type limit problems in this forum..)
    Neither do I. But I can help you.

    Use L'hopital's Rule. If you have a limit that approaches 0, the function is a fraction, and the solution comes out undefined, get the derivative of the numerator, and the derivative of the denominator, and you should be able to solve it. I'll do number one:

    lim x-->0 x^2/1-cosx

    becomes:

    lim x-->0 2x/sinx

    becomes:

    lim x-->0 2/cosx

    becomes:

    2/1

    and you can repeat it as much as you like as long as the limit approaches zero.
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  4. #4
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    Quote Originally Posted by kdl00 View Post
    I'm stuck on getting the limits for sin/cos problems... mind helping me solve these problems?

    1. lim x-->0 x^2/1-cosx
    2. lim x-->0 x/sin3x
    3. lim x-->0 sin5x/sin2x

    (I don't know how to type limit problems in this forum..)
    1. Either use l'Hopital's Rule or substitute the Maclaurin series for cos x and simplify. Or use a useful trig limit found here: World Web Math: Useful Trig Limits

    2. Note that it can be writen as \left(\frac{1}{3}\right) \frac{(3x)}{\sin (3x)}. Now use a useful trig limit.

    3. Note that it can be written as \left( \frac{5}{2}\right) \frac{\sin (5x)}{(5x)} \cdot \frac{(2x)}{\sin (2x)}. Now use a useful trig limit.
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