How to solve this limit

**terrorsquid** · August 24th 2011, 05:05 AM

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

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)*

**anonimnystefy** · August 24th 2011, 05:20 AM

hi terrorsquid

did you try using the l'Hospital's rule? L'Hospital's Rule -- from Wolfram MathWorld

**Prove It** · August 24th 2011, 05:21 AM

Originally Posted by terrorsquid

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

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.

**terrorsquid** · August 24th 2011, 05:24 AM

Originally Posted by anonimnystefy

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.

**Prove It** · August 24th 2011, 05:27 AM

Originally Posted by terrorsquid

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.

**terrorsquid** · August 24th 2011, 05:40 AM

Originally Posted by Prove It

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?

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

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

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

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

$3sec(x)-4tan(x)sin^2(x)$ ?

**Plato** · August 24th 2011, 05:51 AM

Originally Posted by terrorsquid

Is this what you mean?

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

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

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

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

$3sec(x)-4tan(x)sin^2(x)$ ?

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

**terrorsquid** · August 24th 2011, 05:54 AM

Originally Posted by Plato

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

Ah ye, the problem is solved as soon as the sin(x) is gone. I didn't need to keep going.

**CaptainBlack** · August 24th 2011, 06:39 AM

Originally Posted by terrorsquid

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

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

**HallsofIvy** · August 24th 2011, 09:43 AM

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 $\lim_{x\to 0}cos(x)= 1$ .

**terrorsquid** · August 24th 2011, 05:25 PM

Originally Posted by HallsofIvy

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.

**CaptainBlack** · August 24th 2011, 06:59 PM

Originally Posted by terrorsquid

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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