lim t→0 tan 12t sin 4t

Here's another way to do this

tan 12t / sin4t    =

(sin12t/cos12t) / sin4t = (divide numerator and denominator by t)

(sin12t/t) /[(sin4t/t)(cos12t)] = (multiply the numerator by 12/12  and the denominator by 4/4)

(12sin12t/12t) / [ (4sin4t/4t)(cos12t)]

Now

lim t → 0   (12sin12t/12t)   = 12   and

lim t → 0   (4sin4t/4t)  = 4    and

lim t → 0  (cos12t)  = 1     so

12/(4 * 1 )  = 12/4   = 3

$$\\\displaystyle\lim_{t\rightarrow 0}\: \;\frac{tan12t}{sin4t}\\\\\\ =\displaystyle\lim_{t\rightarrow 0}\: \;\frac{sin12t}{cos12t\;sin4t}\\\\\\ =\displaystyle\lim_{t\rightarrow 0}\: \;\frac{sin4tcos8t+cos4tsin8t}{cos12t\;sin4t}\\\\\\ =\displaystyle\lim_{t\rightarrow 0}\: \;\frac{sin4tcos8t+cos4t*2sin4tcos4t}{cos12t\;sin4t}\\\\\\ =\displaystyle\lim_{t\rightarrow 0}\: \;\frac{cos8t+cos4t*2cos4t}{cos12t}\\\\\\ =\frac{cos0+cos0*2cos0}{cos0}\\\\\\ =\frac{1+1*2*1}{1}\\\\ =3$$

Yes that works Chris - it is quite neat too.   I like it    

It is easier than my solution - I always do things the hard way - It is my trade mark so I can't stop now.

Can I ?  Would anyone like me to do it in LaTex so that it is easier to follow?

Thanks, Melody......