# Evaluate the limit Squeeze Theorem Perhaps?

## 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.

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
Homework Helper
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!)

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Ray Vickson