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Math Help - Trig Limit Help neded

  1. #1
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    Trig Limit Help neded

    Hello!

    I've been having trouble solving these 2 trig limits, any help would be appreciated.

    find the limit as x -> 0
    1) sin7x / sin4x

    2) [sin (cos x)] / sec x

    I need to re-arrange both these equations so that 0 can be subbed into x and not get 0 /0. For 1) i multiplied the equation by 7x over 7x and re-arranged it to (sin7x) / 7x * (7x) / sin4x. The sin7x / 7x becomes 1 because of the trig limit properties, but what do I do with 7x / sin4x?

    I don't even know how to start 2

    For those of you that don't know the 2 trig limit properties of the top of your head, here they are:
    limit as x -> 0 of sin x / x = 1
    limit as x -> 0 of cos x - 1 / x = 0

    Thanks for any help
    -Jackson
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  2. #2
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    \frac{{\sin \left( {7x} \right)}}{{\sin \left( {4x} \right)}} = \frac{{\left( {1/x} \right)}}{{\left( {1/x} \right)}}\left( {\frac{{\sin \left( {7x} \right)}}{{\sin \left( {4x} \right)}}} \right) = \frac{{\frac{{7\sin (7x)}}{{7x}}}}{{\frac{{4\sin (4x)}}{{4x}}}}

    P.S. #2 is just \sin (1) . WHY?
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  3. #3
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    Quote Originally Posted by TheJacksonater View Post
    ...
    2) [sin (cos x)] / sec x

    ...
    \frac{\sin(\cos(x))}{\sec(x)}= \frac{\sin(\cos(x))}{\frac1{\cos(x)}}

    And therefore

    \lim_{x \rightarrow 0}\left(\dfrac{\sin(\cos(x))}{\dfrac1{\cos(x)}}\ri  ght) = \sin(1)
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  4. #4
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    Thanks, wow those were easier then I thought now that I see the solutions. Makes me feel pretty stupid
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