trignometric annoyances

[rakuenso]

lim (sin(2x)cos(3x)sin(4x))/(x*cos(5x)sin(6x))

x->0

 

I think I'm missing one key trigonometric property to solve this... it seems too tedious to do all by double angle formulas

[Ducky Havok]

I'm sure there is an easier way to do this, but for now I'll resort to L'Hospital's Rule, which says that if the limit ends up being a indeterminate form then the limit is equal to the limit of the derivative of the top over the derivative of the bottom (Please don't quote me on this, I'm working from memory so that might not be right word for word.)

 

After using the rule once, it ends up being

[math]\lim_{x\to0}\frac{2\cos{2x}\cos{3x}\sin{4x}-3\sin{2x}\sin{3x}\sin{4x}+4\sin{2x}\cos{3x}\cos{4x}}{\cos{5x}\sin{6x}+6x\cos{5x}\cos{6x}-5x\sin{5x}\sin{6x}}[/math]

 

This is still an indeterminate, so use the rule again. For the sake of my typing, if it ends up having a sin(nx) then I just left it out because it'd go to zero, and it'd just be adding and subtracting 0.

 

[math]\lim_{x\to0}\frac{8\cos{2x}\cos{3x}\cos{4x}+8\cos{2x}\cos{3x}\cos{4x}}{6\cos{5x}\cos{6x}+6\cos{5x}\cos{6x}}[/math]

 

Plug in 0 and get (8+8)/(6+6), 16/12, 4/3.

I'm assuming that since it's toward the beginning of the school year you haven't learned L'Hospital's rule yet, so now you're a step ahead of the other once you get there . And if that's still confusing I'm sorry, I'm sure someone will come along later and show a much easier way of doing the problem that I simply overcomplicated.

[rakuenso]

ugh... going from your first diagram to your second diagram requires 6 triple chain rules?

[qrasy]

well, I think the limit can be seen as :

(lim(x->0) sin(2x)/x)*(lim(x->0) cos(3x)/cos(5x))*(lim(x->0) sin(4x)/sin(6x))

 

2*1*4/6

 

4/3

 

to see whether this is correct, try to substitute small x such as 0.001 in calculator and see if the result is close.

[TD]

That is correct, a limit of a product may be calculated as a product of limits.