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Math Help - trig and limits problem

  1. #1
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    phoenix,az
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    trig and limits problem

    hi,

    using sin(θ)/θ x->0 = 1 explain how
    a.) lim (sin2x)/5x = 2/5 as x ->0
    b.)lim (cos h -1)/h = 0 as h->0 using the half angle formula cos h = 1 - 2 sin^ * (h/2)

    since a. is an identity do does the substitution rule apply?

    in b. i can set it up:
    lim -((2sin^2 )*(h/2)/h)

    im not sure what to do or how to solve this limit. my algebra is a little weak so can you explain it to me in simple terms?? any response would be great!
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  2. #2
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    Crna Gora
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    Re: trig and limits problem

    Quote Originally Posted by kingargon View Post
    hi,

    using sin(θ)/θ x->0 = 1 explain how
    a.) lim (sin2x)/5x = 2/5 as x ->0
    b.)lim (cos h -1)/h = 0 as h->0 using the half angle formula cos h = 1 - 2 sin^ * (h/2)

    since a. is an identity do does the substitution rule apply?

    in b. i can set it up:
    lim -((2sin^2 )*(h/2)/h)

    im not sure what to do or how to solve this limit. my algebra is a little weak so can you explain it to me in simple terms?? any response would be great!
    a)

    substitute 2x=t \Rightarrow 5x=\frac{5t}{2}

    also : x \rightarrow 0 \Rightarrow t \rightarrow 0

    Hence :

    L=\displaystile \lim_{t \to 0} \frac{\sin t}{\frac{5t}{2}}=\displaystile \lim_{t \to 0}\left (\frac{2}{5} \cdot \frac{\sin t}{t}\right)=\frac{2}{5} \cdot \displaystile \lim_{t \to 0} \frac{\sin t}{t} =\frac{2}{5}

    b)

    Hint :

    \displaystile \lim_{t \to 0} \frac{\sin^2 t}{t}=\displaystile \lim_{t \to 0} \sin t \cdot \displaystile \lim_{t \to 0} \frac{\sin t}{t}
    Last edited by princeps; April 4th 2012 at 08:56 PM.
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  3. #3
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    Re: trig and limits problem

    great answer! thank you for the quick response.

    i dont understand some of your steps can you explain how u got the substitution values and your arithmitic to find the limit?
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  4. #4
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    Re: trig and limits problem

    Quote Originally Posted by kingargon View Post
    great answer! thank you for the quick response.
    i dont understand some of your steps can you explain how u got the substitution values and your arithmitic to find the limit?
    \frac{\sin(2x)}{5x}=\left(\frac{2}{5}\right)\left(  \frac{\sin(2x)}{2x}\right)
    Now if x\to 0 then also 2x\to 0
    Last edited by Plato; April 5th 2012 at 01:22 PM.
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