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Math Help - [SOLVED] Trig limit Question

  1. #1
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    [SOLVED] Trig limit Question

    Can someone please confirm that this expression's limit DNE. I'm wasting too much paper to try and find out.

    lim x-->0 (tan3x)/(tan5x)


    Also does anyone have any tips on knowing where to stop for when a trig function has a limit that doesn't exist. I find that i'm just spinning my wheels on a lot of the questions that don't have limits.

    Thanks!
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Noxide View Post
    Can someone please confirm that this expression's limit DNE. I'm wasting too much paper to try and find out.

    lim x-->0 (tan3x)/(tan5x)


    Also does anyone have any tips on knowing where to stop for when a trig function has a limit that doesn't exist. I find that i'm just spinning my wheels on a lot of the questions that don't have limits.

    Thanks!
    It does indeed exist!

    \lim_{x\to0}\frac{\tan(3x)}{\tan(5x)}=\lim_{x\to0}  \frac{\sin(3x)\cos(5x)}{\sin(5x)\cos(3x)} =\left(\lim_{x\to0}\frac{5x\sin(3x)}{3x\sin(5x)}\c  dot\frac{3x}{5x}\right)\left(\lim_{x\to0}\frac{\co  s(5x)}{\cos(3x)}\right) =\left(\lim_{x\to0}\frac{\sin(3x)}{3x}\right)\left  (\lim_{x\to0}\frac{5x}{\sin(5x)}\right)\left(\lim_  {x\to0}\frac{3x}{5x}\right)\cdot 1 =1\cdot1\cdot\frac{3}{5}=\boxed{\frac{3}{5}}

    Alternatively, if you know L'Hopital's Rule, you could just do:

    \lim_{x\to0}\frac{\tan(3x)}{\tan(5x)}=\lim_{x\to0}  \frac{3\sec^2(3x)}{5\sec^2(5x)} = \boxed{\frac{3}{5}}
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  3. #3
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    Thanks.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Noxide View Post
    I don't think I quite understand what you did in the expression following the 2nd equals sign. It would be great it you, or anyone could explain this.
    \lim_{x\to0}\frac{\sin(3x)\cos(5x)}{\sin(5x)\cos(3  x)}=\lim_{x\to0}\left(\frac{3x\cdot5x\cdot\sin(3x)  }{3x\cdot5x\cdot\sin(5x)}\cdot\frac{\cos(5x)}{\cos  (3x)}\right) =\left(\lim_{x\to0}\frac{5x\cdot\sin(3x)}{3x\cdot\  sin(5x)}\cdot\frac{3x}{5x}\right)\left(\lim_{x\to0  }\frac{\cos(5x)}{\cos(3x)}\right)

    by the Limit product rule: \lim_{x\to a}f(x)g(x)=\left(\lim_{x\to a}f(x)\right)\left(\lim_{x\to a}g(x)\right) assuming both limits exist.
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  5. #5
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    Yes, thanks again. The answer came to me a few seconds after I hit the post button. Another question did come up. How did you simplify

    limx-->0 (cos(5x)/cos(3x))

    and am I correct in saying that it became the expression limx--> 3x/5x?
    If so, how did it become that?
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Noxide View Post
    Yes, thanks again. The answer came to me a few seconds after I hit the post button. Another question did come up. How did you simplify

    limx-->0 (cos(5x)/cos(3x))

    and am I correct in saying that it became the expression limx--> 3x/5x?
    If so, how did it become that?

    \lim_{x\to0}\frac{\cos(5x)}{\cos(3x)} became 1. You can get that just by plugging in zero. (Remember that \cos(0)=1.)

    The \lim_{x\to0}\frac{3x}{5x} came about as a result of my second usage of the limit product rule. I broke up

    \lim_{x\to0}\frac{5x\sin(3x)}{3x\sin(5x)}\cdot\fra  c{3x}{5x}

    into

    \left(\lim_{x\to0}\frac{\sin(3x)}{3x}\right)\left(  \lim_{x\to0}\frac{5x}{\sin(5x)}\right)\left(\lim_{  x\to0}\frac{3x}{5x}\right)
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  7. #7
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    Thank you. I really appreciate your explanations.
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  8. #8
    Super Member redsoxfan325's Avatar
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    You're welcome.
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