Question

Evaluate on lim x->0 (1-cosx) / x sinx

Answer 1

Hint:

L'Hopital's rule works wonders here.

Answer 2

wat's that

Answer 3

1/2

Answer 4

oh ok

Answer 5

want me to show you how or no?

Answer 6

k show me
thxs

Answer 7

oh ok so numerator can be derived separately from denominator. am i correct??

Answer 8

Kinda. Currently if you plug in 0, you'll get:

lim x->0 (1-cosx) / x sinx
lim x->0 (1-cos0) / 0* sin0
lim x->0 (1-1) / 0
lim x->0 0/0

This is your cue to use l'hopital's rule. So we differentiate the top and bottom first.

Answer 9

d/dx (1-cosx) = sinx (this will be your new numerator)
d/dx (xsinx) = (using product rule) sinx + xcosx (This is your new denominator)

Answer 10

Tips:d/dx means to differentiate (What some people call "deriving", even though that's not the right word. :P)

And make SURE you use product rule for xsinx. Don't assume that the x turns into 1 and that sinx just turns into cosx, resulting in 1cosx. That is WRONG.

Anyhow, the correct formula should now read:

sinx/(sinx+cosx)
Plug in 0. You get 0/(1+0).
Or, 0/1, or 0.

I'm actually a lil worried that lagrange got a different answer. Did I do this correctly, lagrange? We didn't do much l'hopital in my class.

Answer 11

but

Answer 12

isnt 1- sin2x = cosx

Answer 13

your on the write track

Answer 14

It could be, but we didn't use that anywhere.

Answer 15

i will post the solution okay

Answer 16

Sure. I realized where I went wrong.

"Anyhow, the correct formula should now read:
sinx/(sinx+cosx) Plug in 0. You get 0/(1+0). Or, 0/1, or 0. "

That part is waaayyyy wrong. It SHOULD say:

"Anyhow, the correct formula should now read:

sinx/(sinx+Xcosx) Plug in 0.

You get 0/(0+0), which gives you 0/0 again.

But that's ok. Because we're going to use l'hopital's rule AGAIN. :P

Answer 17

Here it is:By L'Hospital's Rule
(It is necessary that the numerator and denominator both tend to 0 or infinity simultaneously i.e a 0/0 or inf./inf. form to apply this rule)
this is a 0/0 form
differentiating num. and den.
sin x/(xcosx + sinx)
it is still 0/0 form
So,L'Hospital rule can be apllied once more
differentiating numerator and denominator again
cosx/(-x sin x+ cos x + cos x)
as limit approaches zero,this tends to 1/2=0.5 which is the answer.

By series expansion
When x is very small
sin x ~x, cos x ~1-x*x/2
substituting these we get the limit to be 1/2

Answer 18

d/dx (sinx/(sinx+Xcosx)) =
cosx/(cosx + [cosx-sinx])

Remember: product rule on xcosx to get the stuff in the brackets.

Now, we plug in zero again:
1/(1+1-0)
1/2

Beautiful.

Answer 19

Oh, whoops. Lagrange beat me to it. :P
Both of us did the exact same thing; if one of us is confusing (probably mine is more confusing), use the other one to make sense of it. Remember: we used the exact same thing.