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

[Denomination]

Find the limit value of (3sin6x-2tan4x)/2x as x goes to 0.

I understand that 2tan4x is supposed to be turned into -2(sin4x/cos4x) but I'm having trouble recognizing the next step. I'm thinking that I should set 3sin6x with a common denominator to the -2(sin4x/cos4x).

 

[daon]

Re: Trig limit

Find the value of (3sin6x-2tan4x)/2x as the limit goes to 0. I understand that 2tan4x is supposed to be turned into -2(sin4x/cos4x) but I'm having trouble recognizing the next step. I'm thinking that I should set 3sin6x with a common denominator to the -2(sin4x/cos4x).

\(\displaystyle \L \frac{3sin6x-2tan4x}{2x} = 9 (\frac{sin6x}{6x}) - (\frac{tan4x}{x}) = 9( \frac{sin6x}{6})- 4(\frac{sin4x}{4x}) \frac{1}{cos4x}\)

Hope that helps.

 

[Denomination]

Confused

In a way it did but I also got a little confused. Because we learned a rule that (sinx/x)=1 so I redid the problem and this is what I got.

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

and I know that (sin6x/6x) goes to one so

9(1)-(2tan4x/2x) is where I am stuck now.

 

[daon]

Yes, that is correct. Now transform (2tan4x)(2x) as I did, into sin(4x)/[xcos(4x)] = 4[(sin(4x))/(4x)][1/cos4x]. Now, look at my decomposition...

Take limit as x goes to zero: sin(6x)/(6x) = 1, sin(4x)/(4x) = 1 and 1/cos(4x) = 1/cos0 = 1.

So we have 9(1)-4(1)(1) = 5.

 

[Denomination]

Thank YOU!

Thanks for coming back and breaking down your explanation for me, I greatly appreciate it. I now see what had to be done in order to further simplify the problem. One last question, it's nothing big but how did you get the cos(4x) in 1/cos(4x) to equal cos0? I understand where you go from there but I am having trouble seeing how the cos(4x) goes to cos0.

Thanks again!

 

[daon]

\(\displaystyle Lim_{x->0} \,\, cos(4x) = cos(4*0) = cos(0)\)

 

[Denomination]

Got it.

I see it now.

Thanks,
Nick V.