Need help: limx->0 sin4x/sin6x

[user2773]

lim x->0 sin4x/sin6x=?
also
Given y=sin2x-2sinx, find dy/dx=0
also
Given f(x)=xg(x^2), find f'' in terms of g,g',g''
Any help would be much appreciated!

 

[Deleted member 4993]

lim x->0 sin4x/sin6x=?

lim x->0 sin4x/sin6x = lim x->0 [(sin4x/4x)/{sin6x/(6x)} * {4x/(6x}]

Now continue...

also
Given y=sin2x-2sinx, find dy/dx=0

Please show your work and indicate excatly where you are stuck

also

Given f(x)=xg(x^2), find f'' in terms of g,g',g''

Please show your work and indicate excatly where you are stuck

Any help would be much appreciated!

 

[komizuki]

1:
L'Hospital rule:

if: lim(x->0) Sin(4x)/Sin(6x) = 0/0 (generally f(x)/g(x) = 0/0) then do f'(x)/g'(x) and continue until you get a result. here f'(x)/g'(x) = 4Cos(4x)/(6Cos(6x) limit x-> makes Cos -> 1 leaves 4/6 = 2/3

2:
dy/dx = 2(Cos[2x]-Cos[x]) = 0, that leads to x=0; x=2Pi/3; x=4Pi/3 for x in [0,2Pi) there is a 2Pi symmetry of course so solution will reappear periodically.

3:
f(x)=xg(x^2) => df(x)/dx = g(x^2) + 2x^2g'(x^2) (the derivative of g(x^2) gives 2xg'(x^2))
=> d^2f(x)/d^2x = 6x g'(x^2) + 4x^3 g''(x^2)