Question

Limits,

http://screencast.com/t/ZpJTBcGF

Answer 1

they used L'Hopital's Rule at that point.

Answer 2

tha'ts like saying to me this http://screencast.com/t/FPl78QqA is this http://screencast.com/t/5ssvCk22l3

Answer 3

not quite. in the original limit, you're comparing two functions but they both go to zero. What L'Hopital's rule says is that if you compare the limit of the rates that the two function go there, it's the same as the limit of the functions themselves.

What you're really looking at is which one gets to zero faster, and comparatively how much faster.
in your case. the bottom goes to 0 infinitely faster than the top so the whole thing goes to -infinity (because of the sign on the top).

Answer 4

Can you explain me how http://screencast.com/t/FPl78QqA became http://screencast.com/t/5ssvCk22l3

Answer 5

yes
\[\frac{ d }{ dx }\sin^2 x = \frac{ d }{ dx }(\sin x)^2 = 2(\sin x)^1\cdot \frac{ d }{ dx }\sin x=2(\sin x)^1\cdot \cos x = 2 \sin x \cos x\]

Answer 6

thanks

Answer 7

you can also break up the original limit to see what I was saying in my earlier post...
\[\lim_{x \rightarrow 0^+}\frac{ -x }{ \sin^2 x }= \lim_{x \rightarrow 0^+}\frac{ x }{ \sin x }\cdot\frac{ -1 }{ \sin x }=\lim_{x \rightarrow 0^+}\frac{ x }{ \sin x }\cdot \lim_{x \rightarrow 0^+}\frac{ -1 }{ \sin x } = 1\cdot -\infty\]
\[\lim_{x \rightarrow 0^+}\frac{ x }{ \sin x }=\lim_{x \rightarrow 0^+}\frac{ 1 }{ \frac{\sin x}{x} }=\frac{1}{\lim_{x \rightarrow 0^+}\frac{\sin x }{ x }}=1\]