# Limit of a trig functions

• Jan 19th 2009, 08:28 PM
Arez
Limit of a trig functions
hi, can someone show me why $\displaystyle sin(3x)/2x$ = 3/2? is it because they multiplied $\displaystyle sinx/x$ by 3/2? if so why does 3sinx = sin3x??
• Jan 19th 2009, 08:43 PM
mr fantastic
Quote:

Originally Posted by Arez
hi, can someone show me why $\displaystyle sin(3x)/2x$ = 3/2? is it because they multiplied $\displaystyle sinx/x$ by 3/2? if so why does 3sinx = sin3x??

$\displaystyle \lim_{x \rightarrow 0}\frac{\sin (3x)}{2x} = \lim_{x \rightarrow 0} \left( \frac{3}{2} \cdot \frac{\sin (3x)}{3x} \right) = \frac{3}{2} \cdot \lim_{t \rightarrow 0}\frac{\sin t}{t}$ where $\displaystyle t = 3x$ ....
• Jan 19th 2009, 09:37 PM
Arez
on that note then, how would I do $\displaystyle sin^2(3t)/2t$? the squared is throwing me off..
• Jan 19th 2009, 09:42 PM
chiph588@
First off I think a new thread is in order but...

$\displaystyle \lim_{x \to 0} \frac{\sin^2 (3x)}{2x} = \left( \lim_{x \to 0} \sin(3x) \right) \cdot \left( \lim_{x \to 0} \frac{\sin(3x)}{2x} \right) \; \; ...$