Question

Could someone please help me on this question:

Find the value of the constant k if

lim x->0 [(sin^-1(kx)) / (2x)] = 3

Please show me the steps you take, the answer is k=6.

Thanks!

Answer 1

L'Hopital's rule is your friend here.

Answer 2

Ohh i see, thanks I'll try that.

Answer 3

What would be the derivative of sin^-1?

Answer 4

Actually nvm

Answer 5

Any chance you could show me the first few steps, apparently I keep getting it wrong,

Answer 6

what's the derivative of arcsin(x)?

Answer 7

(1)/(1+x^2)?

Answer 8

No, that's arctan.

Answer 9

Oh, 1/(sqroot(1-x^2)?

Answer 10

yes. hence using the chain rule, what's the derivative of arcsin(kx)?

Answer 11

Hmm, I'm confused on the two letters, could you tell me this step?

Answer 12

k is a constant.

Answer 13

what's the derivative of cos(kx)?

Answer 14

-sin(kx) . (k)

Answer 15

yes. Now then, using the chain rule as you just did with cos(kx),
what is the derivative of arcsin(kx)?

Answer 16

1/(sqroot(1-x^2) . (k)

Answer 17

almost.

Answer 18

1-(sqroot(1- k^2.x^2) . k?

Answer 19

yes

Answer 20

Hence applying l'Hopital's rule to your original lint, it is equal to the limit of
what/what ?

Answer 21

...original LIMIT, it is equal to...

Answer 22

Hmm, Yes is should be equal to I think

Answer 23

So then lim x-> 0 I'll get k/2 when substituting 0 and then it will be k/2 = 3 which will equal to k = 6

Answer 24

yes

Answer 25

Oh my goodness, thanks for you help! I understand this crap now! haha :D