Does Sin(3x) = 3sinx have a provable limit at theta = 0?

[Wooh]

This is a bizarre little proof that is probably wrong, but I want to know where
\lim_{\theta\rightarrow0}\frac{\sin{3\theta}}{\theta} = 3
This is provable quite easily, I don't think I need to do that atm...
well, we know that
\lim_{\theta \rightarrow 0} \frac{\sin{\theta}}{\theta} = 1
So that would imply that
3\lim_{\theta \rightarrow 0} \frac{\sin{\theta}}{\theta} = 3
thus implying that
\lim_{\theta \rightarrow 0} \frac{3\sin{\theta}}{\theta} = 3
By substitution
\lim_{\theta \rightarrow 0} \frac{\sin{3\theta}}{\theta} = \lim_{\theta \rightarrow 0} \frac{3\sin{\theta}}{\theta}
(iffy step)
If you drop the limit on both sides...
\frac{\sin{3\theta}}{\theta}=\frac{3\sin{\theta}}{\theta}
Which oh so quickly becomes
\sin{3\theta}=3\sin{\theta}

 

[quantumdude]

\lim_{\theta \rightarrow 0} \frac{\sin{3\theta}}{\theta} = \lim_{\theta \rightarrow 0} \frac{3\sin{\theta}}{\theta}
(iffy step)
If you drop the limit on both sides...
\frac{\sin{3\theta}}{\theta}=\frac{3\sin{\theta}}{\theta}
Which oh so quickly becomes
\sin{3\theta}=3\sin{\theta}

Yes, that step is "iffy". With it, I can "prove" that sin(3θ) equals any function with the same limit as θ approaches zero. The step is invalid for the same reason that it is invalid to conclude that sin(3θ)=sin(θ) just because those two functions happen to be equal at θ=0.

edit: fixed quote bracket

 

[Wooh]

Ok, cool, I can accept that. :)

 

[cookiemonster]

Sure, for certain values of theta.

cookiemonster

Edit: What Tom said.

 

[Wooh]

Sure, for certain values of theta.

cookiemonster

Edit: What Tom said.

smartass

 

[cookiemonster]

Hey now, I was just saying the same thing Tom was. ;)

And getting to be a smartass is a privilege of being young!

cookiemonster