Evaluate the limit Squeeze Theorem Perhaps?

[crohozen]

Homework Statement

Evaluate the limit, if it exists:

lim$(\frac{sin x}{4x} * \frac{5-5 cos 3x}{2})$
x→0

The Attempt at a Solution

I understand that $\frac{sin(x)}{4x}* \frac{4}{4} =\frac{1}{4}$ but I don't know
what to do next because the 5-5cos3x/2 trips me up. I'm not seeing anything that I can do to it, so I'm thinking the Squeeze Theorem. Any help progressing further would be appreciated.

 

[process91]

Start in pieces. Take the constants to the outside, and you're left with two terms.

$$\lim_{x\rightarrow 0} \left (\frac{\sin x}{4x}\cdot\frac{5-5\cos 3x}{2} \right) = \frac{5}{8}\lim_{x\rightarrow 0} \left ( \frac{\sin x}{x} - \frac{\sin x \cos 3x}{x} \right)$$

See if you can take it from there.

 

[HallsofIvy]

I wouldn't divide it up quite the way process91 does. Look at
$$\frac{5}{8}\frac{sin x}{x}(1- cos(3x))$$
and it should be obvious.

(If there had been an $x^2$ in the denominator rather than x, I would have made it
$$\frac{5}{8}\frac{sin(x)}{x}(3)\frac{1- cos(3x)}{3x}$$
and it would be a bit more interesting!)

 

[Ray Vickson]

You can use the fact that lim(f(x)*g(x)) = (lim f(x)) * (lim g(x)), if both limits on the right exist.

RGV