What Happens to the Limit of x^2sin(x)/(1-cos^2(x)) as x Approaches 0?

[travwg33]

I seem to find a problem with the limit
lim(x->0)

(x^2sin(x))
-----------
1-cos^2(x)

using the limit as x approaches 0 of x over sine of x equals 1 we can simplify as such
(x) * (x) * sin(x)
----- ------
sin(x) sin(x)
and this would equal 0

but what if we split it like:
x^2 * sin(x)
----- -------
sin(x) sin(x)
then you get
x * x
----
sin(x)
which equals x and then infinity why is this different did i screw up on the second split?

 

[Dick]

x*x/sin(x)=(x/sin(x))*x. The first factor approaches 1. The second factor approaches 0. The limit is zero. Same conclusion. I don't know where you see the 'infinity'.

 

[PBJinx]

(x) * (x) * sin(x)
----- ------
sin(x) sin(x)

is equivalent to

x * x
----
sin(x)

at this point in the second one, we would get zero over zero

using hopital's rule

we take the derivative of the top and then the bottom

giving us

2X/cos(x)

which the limit can be taken of easily

I think that is what you were asking for. right?

 

[travwg33]

but what if we split the second function like the first into x over sin(x) times x and then use the limit rule i first referred to and then be just left with the x is there something wrong with doing that? It seems like there must bc it gives a whole different answer but why is doing that wrong

 

[Dick]

but what if we split the second function like the first into x over sin(x) times x and then use the limit rule i first referred to and then be just left with the x is there something wrong with doing that? It seems like there must bc it gives a whole different answer but why is doing that wrong

I really don't see why you think it gives a whole different answer. It's still 0, either way. Why do you think it isn't?

 

[travwg33]

ok so with this split
x * x
----
sin(x)
without using hopital the
x term equals 1 and ur left with an x
----
sin(x)

 

[Dick]

ok so with this split
x * x
----
sin(x)
without using hopital the
x term equals 1 and ur left with an x
----
sin(x)

Right. And x approaches 0, yes? The limit is still 0.

 

[travwg33]

hhahaha wow now i feel like a tard yeah it does idk what my prob was there for some reason it was in my head that it was alim to infinity at the end of the problem