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Math Help - limits of trigonometric fxns...?

  1. #1
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    limits of trigonometric fxns...?

    okay so i'm confused as to HOW to go about finding the limits of these trigonometric fxns....


    lim x-->0 tan(2x) / sin(3x)cos(4x)


    and


    lim x-->0 sin(x^2 - 9) / x-3
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by veezey11 View Post
    okay so i'm confused as to HOW to go about finding the limits of these trigonometric fxns....


    lim x-->0 tan(2x) / sin(3x)cos(4x)
    We should use the fact that \lim_{u\to0}\frac{\tan u}{u}=1 and \lim_{u\to0}\frac{\sin u}{u}=1

    Thus, \lim_{x\to 0}\frac{\tan(2x)}{\sin(3x)\cos(4x)}=\lim_{x\to0}\f  rac{\tan(2x)}{\sin(3x)}\cdot\underbrace{\lim_{x\to  0}\frac{1}{\cos(4x)}}_1=\lim_{x\to0}\frac{\display  style\frac{\tan(2x)}{x}}{\displaystyle\frac{\sin(3  x)}{x}}=\lim_{x\to0}\frac{\displaystyle\frac{\tan(  2x)}{x}\cdot{\color{red}\frac{2}{2}}}{\displaystyl  e\frac{\sin(3x)}{x}\cdot{\color{red}\frac{3}{3}}} =\lim_{x\to0}\frac{2\displaystyle\frac{\tan(2x)}{2  x}}{3\displaystyle\frac{\sin(3x)}{3x}}=\frac{2}{3}  \frac{\displaystyle\lim_{x\to0}\frac{\tan(2x)}{2x}  }{\displaystyle\lim_{x\to0}\frac{\sin(3x)}{3x}}=\f  rac{2}{3}\cdot\frac{1}{1}=\color{red}\boxed{\frac{  2}{3}}

    Does this make sense?

    --Chris
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    uhhhm sortof yeah... i'll take a closer look at it. and see if i can figure it all out. THANK YOU SO MUCH!


    <3
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    Quote Originally Posted by veezey11 View Post
    okay so i'm confused as to HOW to go about finding the limits of these trigonometric fxns....


    lim x-->0 tan(2x) / sin(3x)cos(4x)


    and


    lim x-->0 sin(x^2 - 9) / x-3
    The easier way is with L'Hospital's rule, namely

    \lim_{x \to a} \frac{g(x)}{h(x)} = \lim_{x \to a} \frac{g'(x)}{h'(x)}.


    So \lim_{x \to 0} \frac{\tan{2x}}{\sin{3x}\cos{4x}} = \lim_{x \to 0} \frac{\frac{d}{dx}(\tan{2x})}{\frac{d}{dx}(\sin{3x  }\cos{4x})}

     = \lim_{x \to 0} \frac{\frac{2}{\cos^2{2x}}}{3\cos{3x}\cos{4x} - 4\sin{3x}\sin{4x}}

     =  \lim_{x \to 0} \frac{2}{\cos^2 {2x}[3\cos{3x}\cos{4x} - 4\sin{3x}\sin{4x}]}

    Can you go from here?
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    Quote Originally Posted by veezey11 View Post
    okay so i'm confused as to HOW to go about finding the limits of these trigonometric fxns....


    lim x-->0 tan(2x) / sin(3x)cos(4x)


    and


    lim x-->0 sin(x^2 - 9) / x-3
    For the second, why can't you just use direct substitution?
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