how to evalute trigonometric limits

[EriLynn.]

So i'm just reviewing for a course i will be taking soon, I took a calculus last year, and for some reason i cannot recall, nor find in my text book, how to solve these kinds of questions.
A question like " limit as x approaches 0 of (sin(2x)/3x) ....
I tried changing the sin(2x) into 2sinxcosx... but after that i dont know what to do
could anyone help me out, i just need to know the idea and what kind of steps to follow
thanks a million folks!

 

[galactus]

You do not have to change sin(2x) to 2sin(x)cos(x).

Do this, bring out the 1/3 outside the limit:

\(\displaystyle \frac{1}{3}\lim_{x\to 0}\frac{sin(2x)}{x}\)

Multiply top and bottom by 2:

\(\displaystyle \frac{1}{3}\lim_{x\to 0}\frac{2sin(2x)}{2x}\)

\(\displaystyle \frac{2}{3}\underbrace{\lim_{x\to 0}\frac{sin(2x)}{2x}}_{\text{limit is 1}}\)

\(\displaystyle \frac{2}{3}\cdot 1=\frac{2}{3}\)

 

[BigGlenntheHeavy]

Another way:

\(\displaystyle \lim_{x\to0}\frac{sin(2x)}{3x} \ gives \ the \ indeterminate \ form \ \frac{0}{0}.\)

\(\displaystyle Ergo, \ the \ Marqui \ to \ the \ rescue. \ \frac{D_x[sin(2x)]}{D_x(3x)} \ = \ \frac{2cos(2x)}{3},\)

\(\displaystyle hence \ \lim_{x\to0}\frac{2cos(2x)}{3} \ = \ \frac{2(1)}{3} \ = \ \frac{2}{3}.\)