Finding $\lim_{x\to 0}\frac{\sin 6x}{x}$ Why are we allowed to multiply the numerator and denominator by 6?
Also, when you multiply the numerator by 6 why doesn't the 6 multiply to 36? 
Please share any techniques that can clarify how to evaluate this limit. The concept is confusing me now that I can't just either substitute, factor, or use the conjugate methods. 
 A: QUOTE Why are we allowed to multiply the top and the numerator and denominator by 6? END OF QUOTE
$$
\frac{\sin(6x)}x = \frac{6\sin(6x)}{6x}
$$
What has been been multiplied by $6$ here is just the numerator and the denominator.  It would be equally valid to multiply them both by $13$, thus:
$$
\frac{\sin(6x)}x = \frac{13\sin(6x)}{13x}
$$
but that would not get us where we want to go.  Your phrasing, "the top and the numerator and denominator" makes me wonder if you thought that three things were being multiplied by $6$.
QUOTE Also, when you multiply the numerator by 6 why doesn't the 6 multiply to 36? END OF QUOTE
Because we are nowhere multiplying $6$ by $6$.
A: Notice:
$$\lim_{x\to 0}\frac{\sin(6x)}{x} = \frac{6}{6} \lim_{x\to 0}\frac{\sin(6x)}{x} = \lim_{x\to 0}6\frac{\sin(6x)}{6x}$$
Letting $u=6x$
$$\lim_{x\to 0}\frac{6\sin(6x)}{x} =\lim_{u\to 0} \frac{6\sin(u)}{u}$$
By L'Hospital's rule
$$\lim_{u\to 0} \frac{6\sin(u)}{u} = \lim_{u\to 0} \frac{\frac{d}{du}6\sin(u)}{\frac{d}{du}u} $$$$= \lim_{u\to 0}\frac{6\cos(u)}{1} = 6$$  
A: You're allowed to multiply any expression by $1$ without changing the expression.
But $\dfrac 6 6 = 1$.
(Regarding the substitution $\theta = 6x$, it isn't wrong, but it's an unnecessary step. It doesn't hurt, if it helps your understanding.)
