Finding the limit of trigonometric functions? (1 Viewer)

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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:
31,919
3,891
Do you know this limit?
[tex]\lim_{x \rightarrow 0} \frac{sin(x)}{x} = 1[/tex]
 
yes, but I dont know how to get that point
 
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 :/
 

lanedance

Homework Helper
3,305
2
sounds like extra complication, do you know L'Hopitals rule?

(that said end up in a similar place)
 
L'Hopitals is the way to go. did not try it with this problem but
one of the best "tools" i ever learned
 
31,919
3,891
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 :/
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.
 
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?

edit: nvm someone answered
 

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