Question

Please show me how to solve this limit.
limit of 1-cosx/sin^(2)x as x approaches 0.

Answer 1

This:
\[\lim_{x \rightarrow 0} \frac{1-cosx}{\sin^2x}\]

Answer 2

yeah.

Answer 3

Use L'Hopital's rule

Answer 4

We've just got into the book recently and haven't learned that rule yet.

Answer 5

Hmm, I'm not really sure how you would do it without L'Hopital's rule...

Answer 6

\[\lim_{x \rightarrow 0}\frac{ 1-\cos(x) }{ \sin^2(x) }=\lim_{x \rightarrow 0}\frac{ 1-\cos(x) }{ 1-\cos^2() }=\lim_{x \rightarrow 0}\frac{ 1-\cos(x) }{[1-\cos(x)][1+\cos(x)] }=\]\[\lim_{x \rightarrow 0}\frac{ 1 }{ 1+\cos(x) }=\frac{ 1 }{ 2 }\]

Answer 7

Ohhh. I was doing all kind of things and I didn't think of this way.... Thank you Carlos.

Answer 8

you are welcome!