# Finding the limit of trigonometric functions?

1. Feb 24, 2009

### 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 Im doing this problem right?

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

sin (cosx) / 1/cosx

sin (cosx) cosx

sin (1) 1 = sin 1

Last edited: Feb 24, 2009
2. Feb 24, 2009

### Staff: Mentor

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

3. Feb 24, 2009

### r_swayze

yes, but I dont know how to get that point

4. Feb 24, 2009

### r_swayze

just thought of something but I dont 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 Im off track :/

5. Feb 24, 2009

### lanedance

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

(that said end up in a similar place)

6. Feb 24, 2009

### fball558

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

7. Feb 24, 2009

### Staff: Mentor

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.

8. Feb 24, 2009

### r_swayze

no we havent learned L'Hopitals yet, just started adding this week with trig and limits.

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