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Math Help - limit

  1. #1
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    limit

    I have always taken for granted sin, cos, and tan... but now i want to try to understand something about the unit circle which i have neglected through high school... lol. Well, I have a statement that says sin x < x < tan x. Can someone please bring me through why this is the case geometrically. Also, i am able to prove lim(x->0) sin(x)/(x) = 1 from this but just wondering what the proof for lim(x->0) sin(kx)/(x) = k looks like.
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  2. #2
    Super Member wingless's Avatar
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    \lim_{x\to 0}\frac{\sin kx}{x}

    \lim_{x\to 0}\frac{\sin kx}{x}\cdot \frac{k}{k}

    \lim_{x\to 0}\frac{\sin kx}{kx}\cdot k

    When x approaches zero, kx will approach zero too. So we can write it as,

    \lim_{kx\to 0}\underbrace{\frac{\sin kx}{kx}}_{\text{This makes 1}}\cdot k

    1\cdot k = k
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  3. #3
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    Quote Originally Posted by ah-bee View Post
    I have always taken for granted sin, cos, and tan... but now i want to try to understand something about the unit circle which i have neglected through high school... lol. Well, I have a statement that says sin x < x < tan x. Can someone please bring me through why this is the case geometrically.
    Try this PlanetMath: comparison of $\sin þeta$ and $þeta$ near $þeta = 0$

    Quote Originally Posted by ah-bee View Post
    Also, i am able to prove lim(x->0) sin(x)/(x) = 1 from this but just wondering what the proof for lim(x->0) sin(kx)/(x) = k looks like.
    Hint: \lim_{x \to 0} \frac{\sin(kx)}{x} =  k\lim_{x \to 0} \frac{\sin(kx)}{kx}
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