Help Me Please!!!!!!!!?

Question

Find Lim(2sin2x/2x)(3x/3sin3x)
x>0

Answer 1

i never took geometry i have no idea what you are asking me ask your parents, friends or teachers.

Answer 2

Yeah, using L'Hopitals rule, (0/0) case, you need to take the deriv. of the top over the deriv. of the bottom. You should get the following:

(3(2sin2x) + 3x(4cos2x)) / (2(3sin3x) + 2x(9cos3x))

Plugging in 0 for x here also gives us a 0 / 0 case, so we need to take the deriv. of the top and the deriv. of the bottom again to get the following:

( 12cos2x + 12cos2x - 3x(8sin2x)) /
(18sin3x + 18cos3x - 2x(27sin3x))

Now, plugging in 0 for x should give you 24 / 36 which simplifies to 2/3.

Answer 3

Lim(2sin2x/2x)(3x/3sin3x)
x --> 0
lim sin 2x/2x = 1 <-- well known limit
x --> 0
lim sin3x/3x = 1 <-- well known limit
x --> 0
Lim(2sin2x/2x)(3x/3sin3x) = 2*1*(1/3)*1 = 2/3
x --> 0

Answer 4

2/3

Try using L'Hopital's Rule

lim as x->0 of sin(a*x)/(a*x) = 0/0 is undefined so you can use L'Hopital's Rule,

you take the derivative of the top and the bottom so it becomes

lim as x->0 of sin(a*x)/(a*x) = lim as x->0 of a*cos(a*x)/a = cos(a*x) and

lim as x->0 of cos(a*x) = 1. Apply this and you should get 2/3.

Answer 5

that will simplify down to

lim sin2x/sin3x
x>0

Answer 6

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Answer 7

Sorry but i this is too hard

Answer 8

sorry not smart enough lol and is that suppose to be english

Answer 9

Sod that!

Answer 10

to hard