1. ## Trigonometry Limits

I have this problem:

lim (x-->1) sin(2x)/3x

I multiplied the whole thing by 2/2 and took a 2/3 out of the limit, so I then have:
2/3 [lim(x-->1) sin(2x)/2x

Which of course is very similar to the sinx/x limit, but the variable is approaching 1, not 0 in this case...

Would it be correct to do this:
Let z = x-1
x-->1, then z-->0
and turn the equation into:
(2/3) [lim(z-->0) sin2(z+1)/2(z+1)
I feel like this isn't going to get me anywhere useful...

Any tips, please? I appreciate it.

2. The function $\displaystyle \frac{\sin(2x)}{3x}$ is continuous at 1 so $\displaystyle \lim_{n \to 1} \frac{\sin(2x)}{3x} = \frac{\sin(2 \cdot 1)}{3 \cdot 1} = \frac{\sin(2)}{3}$.

3. Originally Posted by Giraffro
The function $\displaystyle \frac{\sin(2x)}{3x}$ is continuous at 1 so $\displaystyle \lim_{n \to 1} \frac{\sin(2x)}{3x} = \frac{\sin(2 \cdot 1)}{3 \cdot 1} = \frac{\sin(2)}{3}$.
So no fancy manipulation to make it into that trig limit format? Wow... nice and simple. Thanks Giraffo!