Finding the limit of trigonometric functions?

[r_swayze]

Im having trouble with this problem here:

x-->0 tan6x / sin2x

So far I only have:

(sin6x / cos6x) / (2sinx cosx)

What would be the next step? Multiply the reciprocal?
Also, could somebody tell me if I am doing this problem right?

x--> 0 (sin (cosx)) / secx

sin (cosx) / 1/cosx

sin (cosx) cosx

sin (1) 1 = sin 1

 

[Mark44]

Do you know this limit?
$$\lim_{x \rightarrow 0} \frac{sin(x)}{x} = 1$$

 

[r_swayze]

yes, but I don't know how to get that point

 

[r_swayze]

just thought of something but I don't know if its right. Am I suppose to separate sin6x and sinx and then apply the sin(x)/x = 1 theorem?

(sin6x/1)(1/cos6x)
------------------
(sinx/1) 2cosx

then multiply by 1/x and apply the theorem?

6 / cosx
---------
2 cosx

Does this sound right? I feel like I am off track :/

 

[lanedance]

sounds like extra complication, do you know L'Hopitals rule?

(that said end up in a similar place)

 

[fball558]

L'Hopitals is the way to go. did not try it with this problem but
one of the best "tools" i ever learned

 

[Mark44]

just thought of something but I don't know if its right. Am I suppose to separate sin6x and sinx and then apply the sin(x)/x = 1 theorem?

(sin6x/1)(1/cos6x)
------------------
(sinx/1) 2cosx

then multiply by 1/x and apply the theorem?

6 / cosx
---------
2 cosx

Does this sound right? I feel like I am off track :/

You'd like to get sin(6x)/6x and 2x/sin(2x), so it's just a matter of multiplying by 1 in some form that gets you there. Note that since sin(x)/x approaches 1 as x approaches 0, x/sin(x) has the same limit. Also tan(x)/x approaches 1 as x approaches 0. This is actually easier than L'Hopital's Rule, I think, since there's the possibility of screwing up a derivative there.

 

[r_swayze]

no we haven't learned L'Hopitals yet, just started adding this week with trig and limits.

But did anyone check my second posting after Mark44? Does that look right to you guys?

edit: nvm someone answered