Trig limit (3sin6x/2x)-(2tan4x)/2x as x goes to 0

[Denomination]

I apologize for reposting this problem but I didn't understand the repost that I previously got. I got a little farther this time.

The problem is:

Find the value of

limit (3sin6x/2x)-(2tan4x)/2x

as x goes to 0.

This is what I got

3((3sin6x)/2x) - (2tan4x)/2x (i split the problem into two to try and simplify it)

So then I got

9(sin6x/6x)- (2tan4x/2x)

Since (sin6x/6x)=1 I'm left with

9-(2tan4x/2x)

I know that this has something to do with (sin4x/cos4x)=tan4x but I don't know what to do after I convert the tan4x to that

 

[daon]

It isn't necessary to repost the question. I would have been happy to go through the steps more carefully.

You have
\(\displaystyle \L9-\frac{2tan4x}{2x} = 9-\frac{tan2x}{x} = 9-\frac{\frac{sin4x}{cos4x}}{x} \\ = 9 - \frac{sin4x}{xcos4x} = 9- \frac{sin4x}{x} \frac{1}{cos4x}\)

Now, wouldn't it be nice if we had sin4x/4x instead of sin4x/x? We can change it! If you devide by four and multiply by four, you never change anything and it helps us out a great deal.

So,
\(\displaystyle \L9- \frac{sin4x}{x} \frac{1}{cos4x} = 9- \frac{4sin4x}{4x} \frac{1}{cos4x} = 9- 4\frac{sin4x}{4x} \frac{1}{cos4x}\)

Now,take the limit as x->0:

\(\displaystyle Lim_{x->0} 9- 4\frac{sin4x}{4x} \frac{1}{cos4x} = 9-4(1)(\frac{1}{cos0}) = 9-4(1)(1)=5\)