Question

find the limit of:
Limit(x->0) : (cosx-1)/sinx
All i know is you start by performing the quotient rule, but now im stuck! thanks in advance!

Answer 1

Have you learned `L'Hospital's Rule` at this point?

Answer 2

\[\Large\bf\sf \lim_{x\to0}\frac{\cos x-1}{\sin x}\]
We're getting the indeterminate form:\[\Large\bf\sf \frac{0}{0}\]So we have a good candidate for L'H Rule. But have you learned that method yet?

Answer 3

i have not...not yet anyways

Answer 4

Hmm ok here's another method we can try.
Let's multiply the top and bottom by the `conjugate` of the top.

Answer 5

\[\Large\bf\sf \lim_{x\to0}\frac{\cos x-1}{\sin x}\left(\frac{\cos x+1}{\cos x+1}\right)\quad=\quad \lim_{x\to0}\frac{\cos^2-1}{\sin x(\cos x+1)}\]

Answer 6

If you factor a -1 out of the top, you can apply your square identity.\[\Large\bf\sf \lim_{x\to0}\frac{-(1-\cos^2x)}{\sin x(\cos x+1)}\quad=\quad \lim_{x\to0}\frac{-(\sin^2x)}{\sin x(\cos x+1)}\]

Answer 7

From here we can cancel out some sines, yes?
Any step in there really confusing?

Answer 8

This is something to try and get used to with limits.
You plug the value directly in, if there is a problem, you do some algebra and try to cancel stuff out.

After canceling something out, see if you still have a problem plugging the value directly in.

Answer 9

thank you sooo much!!! this helped tons!!!!!
i understand! wahoo

Answer 10

Oh good c: