Question

Please help!!!!!!
lim as x approaches 0 of (2(sin3x)^3/-x^3)

Answer 1

\[\lim_{x \rightarrow 0}\frac{ 2 Sin ^{3} 3x }{ -x ^{3} }\] is this your problem?

Answer 2

Yes!

Answer 3

do you know L'Hopital's rule?

Answer 4

no I do not

Answer 5

if you apply the limit as it is you will get a form 0/0 .

So whenever you face such a situation where your limit turns to the form 0/0, or infinity / infinity, -infinity/infinity, -inifinty/-infinity, you an apply L'Hopitals rule.

Answer 6

For using L'Hopital's rule you need to take the derivative of the numerator separately and the derivative of the denominator separately. An then apply the limit again.

If you face the same situation as earlier, You do the derivative one more time. Then apply the limit.

You keep on taking the derivative until you reach a determinate limit.

Answer 7

here if you apply limit 0 the whole thing becomes 0/0

Now take the derivative of the numerator ( Apply Chain rule) Seperately.

Take the derivative of denominator separately.

Answer 8

So I take the derivative of each portion separately until I get a constant in the denominator?

Answer 9

yes

Answer 10

in this case

Answer 11

but if you have different thing in the denominator say you had Sin x. If you differentiate that it becomes Cosx and if you apply 0 its value is 1.

Answer 12

Thank you!
I believe you get -- the third derivative of f =\[\frac{ -12\cos(3x) }{ -6 }\]

Answer 13

do I solve from there by plugging in 0 for x?

Answer 14

in the end you get 2 -- yes?

Answer 15

for differentiating the numerator you apply chain rule uv

Answer 16

how do you get just -12 cos 3x

Answer 17

yes but by applying the chain rule - you just multiple the numerator by 1

Answer 18

\[2 \sin ^{3} 3x\] that is the numerator right?

Answer 19

the first derivate of the numerator is: \[6 \cos ^{2} (3x)\]

Answer 20

yes

Answer 21

first derivative is 6Sin²3x . Cos 3x . 3 = 18 Sin²3x .Cos 3x

Answer 22

Sin³ 3x = 3. Sin² 3x . Cos 3x. 3 = 9 Sin² 3x. Cos 3x

you already had a 2 in the front

Answer 23

now you have to do uv method

Answer 24

2nd derivative

18 sin² 3x . Cos 3x

= 18[ 2 Sin 3x Cos 3x. 3 . Cos 3x + Sin² 3x . (- Sin 3x) . 3 ]

= 18 [ 6 . Sin 3x. Cos² 3x - 3 Sin³ 3x]

= 108 Sin 3x. Cos² 3x - 54 Sin³ 3x

Answer 25

3rd derivative

108 [ Cos 3x. 3 . Cos² 3x + Sin 3x . 2 Cos 3x . (-sin 3x) . 3] - 54[ 3Sin²3x . Cos 3x . 3]

= 108 [ 3 Cos³ 3x - 6 Sin² 3x . Cos 3x ] - 486 Sin² 3x . Cos 3x

= 324 Cos³ 3x - 648 Sin² 3x . Cos 3x - 486 Sin² 3x . Cos 3x

= 324 Cos³ 3x - 1134 Sin² x . Cos 3x

Answer 26

now if you apply the limit the second term becomes 0. and the first term becomes 324

the denominator you have -6

so your answer will be 324 / -6 = -54