Understanding Arcsin and L'Hopital: Finding the Limit of (arcsin(2x))/x^3

[idris]

Hey guys, have a questions about L'Hopital and arcsin.

The question is to find the limit of (arcsin(2x))/x^3 as x->0. I can find the limit no problem just by applying L'Hopital, but I am having difficulty proving that it's valid to use L'Hopital with the function, because with f(x)=arcsin(2x), the limit of arcsin(2x) as x->0 is 2x, rather than 0 or infinity as is required. I think I may have to express arcsin(2x) as a ratio and then apply L'Hopital again, but I can't figure out how to do that. Thanks for any help!

 

[malawi_glenn]

that is strange, I thought lim (y ->0) arcsin(y) = 0

 

[idris]

I thought that for very small values of sin or arcsin, sin(x) roughly = x? So that if you take the limit as x->0, it will never be 0, but just the input? This is hard to explain - the function doesn't "get closer" to anything as it approaches 0, it just oscillates around the value inputted - so it can't really be said to have a limit as x->0?

Or is that all totally off-base?

EDIT: Yeah, it is. I just needed to read over my trig notes a bit closer. Thanks malawi for putting me back on track...!

 

[HallsofIvy]

I thought that for very small values of sin or arcsin, sin(x) roughly = x? So that if you take the limit as x->0, it will never be 0, but just the input? This is hard to explain - the function doesn't "get closer" to anything as it approaches 0, it just oscillates around the value inputted - so it can't really be said to have a limit as x->0?

Or is that all totally off-base?

EDIT: Yeah, it is. I just needed to read over my trig notes a bit closer. Thanks malawi for putting me back on track...!

No, the limit, as x--> 0 is NUMBER. In general, if f(x) is a continuous function,
\lim_{x\rightarrowa}f(g(x))= f(\lim_{x\rightarrowa}g(x)).