finding the limit

[nickname]

what is the limit for the following function as x---->0?

sin(2x)/3cosx

Is it 2/3?

I "broke up" the equation:

(1/3) (sinx/x) (x/cosx) = (1/3)(1)(2)=2/3?

Thank you!

 

[galactus]

No, sorry to say, it is not 2/3.

Use the identity sin(2x)=2sin(x)cos(x)

$\displaystyle \lim_{x\to 0}\frac{sin(2x)}{3cos(x)}=\lim_{x\to 0}\frac{2sin(x)cos(x)}{3cos(x)}=\lim_{x\to 0}\frac{2sin(x)}{3}$

Now, see the limit?. What is sin(0)?.

Also, your method was fine except $\displaystyle \lim_{x\to 0}\frac{x}{cos(x)}=0, \;\ \text{not 2}$

That would give you 1/3(1)(0)=0, as required.

 

[nickname]

Thanks for informing me!