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Math Help - How to solve this limit

  1. #1
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    How to solve this limit

    \lim_{x \to 0} ~\frac{sin(6x)}{sin(2x)cos(3x)}

    I can see that the limit of each component is zero but can't seem to find the logic behind the entire function being zero. How can I get it into a form where the denominator is not 0 and the limit of the denominator is not zero? The only manipulation I could see was to change the numerator to 2sin(3x)cos(3x) and cancel the cos(3x), but I am left with the same set of problems.

    I remember seeing rules for division and multiplication of trig functions with different angles like cos(2x)/cos(x) = cos(2x-x). Did I just imagine this? I can't seem to find anything on it; I can only find the standard trig identities (sum difference formulas etc.).

    Thanks.

    edit: cos(3x)*
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  2. #2
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    Re: How to solve this limit

    hi terrorsquid

    did you try using the l'Hospital's rule? L'Hospital's Rule -- from Wolfram MathWorld
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  3. #3
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    Re: How to solve this limit

    Quote Originally Posted by terrorsquid View Post
    \lim_{x \to 0} ~\frac{sin(6x)}{sin(2x)cos(3x)}

    I can see that the limit of each component is zero but can't seem to find the logic behind the entire function being zero. How can I get it into a form where the denominator is not 0 and the limit of the denominator is not zero? The only manipulation I could see was to change the numerator to 2sin(3x)cos(3x) and cancel the cos(3x), but I am left with the same set of problems.

    I remember seeing rules for division and multiplication of trig functions with different angles like cos(2x)/cos(x) = cos(2x-x). Did I just imagine this? I can't seem to find anything on it; I can only find the standard trig identities (sum difference formulas etc.).

    Thanks.

    edit: cos(3x)*
    Yes you did just imagine it.

    Anyway, use the facts that \displaystyle \sin{3x} = 3\sin{x} - 4\sin^3{x} and \displaystyle \sin{2x} = 2\sin{x}\cos{x} and see what you can do from there.
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  4. #4
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    Re: How to solve this limit

    Quote Originally Posted by anonimnystefy View Post
    hi terrorsquid

    did you try using the l'Hospital's rule? L'Hospital's Rule -- from Wolfram MathWorld
    We haven't done L'Hospital's yet; I'll look into it though thanks.
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  5. #5
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    Re: How to solve this limit

    Quote Originally Posted by terrorsquid View Post
    We haven't done L'Hospital's yet; I'll look into it though thanks.
    You don't need L'Hospital's Rule in this case, just do what I said in the above post.
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    Re: How to solve this limit

    Quote Originally Posted by Prove It View Post
    You don't need L'Hospital's Rule in this case, just do what I said in the above post.
    Is this what you mean?

    \lim_{x \to 0} ~\frac{sin(6x)}{sin(2x)cos(3x)}

    \frac{2sin(3x)cos(3x)}{2sin(x)cos(x)cos(3x)}

    \frac{3sin(x)-4sin^3(x)}{sin(x)cos(x)}

    \frac{sin(x)(3-4sin^2(x))}{sin(x)cos(x)}

    3sec(x)-4tan(x)sin^2(x) ?
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  7. #7
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    Re: How to solve this limit

    Quote Originally Posted by terrorsquid View Post
    Is this what you mean?

    \lim_{x \to 0} ~\frac{sin(6x)}{sin(2x)cos(3x)}

    \frac{2sin(3x)cos(3x)}{2sin(x)cos(x)cos(3x)}

    \frac{3sin(x)-4sin^3(x)}{sin(x)cos(x)}

    \frac{sin(x)(3-4sin^2(x))}{sin(x)cos(x)}

    3sec(x)-4tan(x)sin^2(x) ?
    \frac{3sin(x)-4sin^3(x)}{sin(x)cos(x)}=\frac{3-4\sin^2(x)}{\cos(x)}\to 3
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    Re: How to solve this limit

    Quote Originally Posted by Plato View Post
    \frac{3sin(x)-4sin^3(x)}{sin(x)cos(x)}=\frac{3-4\sin^2(x)}{\cos(x)}\to 3
    Ah ye, the problem is solved as soon as the sin(x) is gone. I didn't need to keep going.
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  9. #9
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    Re: How to solve this limit

    Quote Originally Posted by terrorsquid View Post
    \lim_{x \to 0} ~\frac{sin(6x)}{sin(2x)cos(3x)}

    I can see that the limit of each component is zero but can't seem to find the logic behind the entire function being zero. How can I get it into a form where the denominator is not 0 and the limit of the denominator is not zero? The only manipulation I could see was to change the numerator to 2sin(3x)cos(3x) and cancel the cos(3x), but I am left with the same set of problems.

    I remember seeing rules for division and multiplication of trig functions with different angles like cos(2x)/cos(x) = cos(2x-x). Did I just imagine this? I can't seem to find anything on it; I can only find the standard trig identities (sum difference formulas etc.).

    Thanks.

    edit: cos(3x)*
    You should know that for small u that \sin(u)\sim u and that \cos(u)\sim 1

    So as x goes to zero:

    \frac{\sin(6x)}{\sin(2x)\cos(3x)}\sim \frac{6x}{2x}=3

    CB
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  10. #10
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    Re: How to solve this limit

    I would have done it as \frac{sin(6x)}{sin(2x)cos(3x)} = 6\left(\frac{sin(6x)}{6x}\right) \left(\frac{1}{2}\right) \left(\frac{2x}{sin(2x)}\right)\frac{1}{cos(3x)}
    and each of those limits is standard: \lim_{x\to 0}\frac{sin(x)}{x}= 1 and \lim_{x\to 0}cos(x)= 1.
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    Re: How to solve this limit

    Quote Originally Posted by HallsofIvy View Post
    I would have done it as \frac{sin(6x)}{sin(2x)cos(3x)} = 6\left(\frac{sin(6x)}{6x}\right) \left(\frac{1}{2}\right) \left(\frac{2x}{sin(2x)}\right)\frac{1}{cos(3x)}
    and each of those limits is standard: \lim_{x\to 0}\frac{sin(x)}{x}= 1 and \lim_{x\to 0}cos(x)= 1.
    Can you explain this. Are those limits identities or something, because I would have thought getting it into the form sin(x)/x is creating the same situation I was trying to avoid. Also, are you applying a particular method in order to rearrange the function or just thinking abstractly?

    Thanks.
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  12. #12
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    Re: How to solve this limit

    Quote Originally Posted by terrorsquid View Post
    Can you explain this. Are those limits identities or something, because I would have thought getting it into the form sin(x)/x is creating the same situation I was trying to avoid. Also, are you applying a particular method in order to rearrange the function or just thinking abstractly?

    Thanks.
    That:

    \lim_{x \to 0} [\sin(x)/x]=1

    is a standard limit that it is reasonable to expect you to know. It expresses the same idea as \sin(x) \sim x as x \to 0.

    The rearrangement is just a technique to turn the original limit into something using other known limits.

    CB
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