Find the limit as x approaches 0 EQUATION INSIDE

Question

anonymous · Answer

$$\lim_{x \rightarrow 0}\frac{ \sin(2x) }{ \sin(16x) }$$

anonymous · Answer

Thanks in advance guys. Please don't use l'hopitals or anything. I just am suposed to transform is algebraically or say it does not exist and why

anonymous · Answer

I know the answer is 1/8

anonymous · Answer

well try making both the same form. what is the problem here :) clue - sin(2x) expands ;)

anonymous · Answer

and so does sin(nx)

anonymous · Answer

what do you mean by expands? like sinx * sin2?

raden · Answer

actually, i u can use the formula :
lim (x->0) sinAx/sinBx = A/B

anonymous · Answer

or do i use that identity for sin 2x?

anonymous · Answer

oh well i guess I can just really shorten that. i was afriad that wasn't mathmatically "legal" thanks though

anonymous · Answer

so its just be 2/16, then 1/8. Thanks

raden · Answer

yes, it will work.. or u can use L'Hopital

shubhamsrg · Answer

What @RadEn says comes directly from taylor series expansions, you can use that, have you learnt that ?

anonymous · Answer

I haven't but Im thinking if i just show exactly what he said I should be ok

shubhamsrg · Answer

sin y = y - y^3 /3!  + y^5 /5! -y^7 /7! ....

ever learnt stuff like that?

shubhamsrg · Answer

Oh wait, there is an alternative way to reach an answer!

shubhamsrg · Answer

You must be knowing sinx /x = 1 when x->0

dumbcow · Answer

sin(2x) = 2sin(x) cos(x)

shubhamsrg · Answer

x->0 sin(x)/x =1 , you know  that right ?