Franklin wrote: but the answer is likely zero the the area under the curve from 0 to 0, unless n is way bigger than m- Right

when n is just 1 larger than m, the limit does exist and is not 0.

before x reached 0, we can treat x as a small amount. 2x is small too.

then the integral can be approximately treated as the area of a trapezoid, with height x=2x-x,

the left base (sinx)^m/x^n, the right base (sin2x)^m/(2x)^n.

Since x and 2x are very small. the left base can be treated as

(x+o(x))^m/x^n= (both*1/x^{m})= (1+o(x))/x

and the right base

(1+o(x))/2x

Hence the area is

[(1+o(x))/x+(1+o(x))/2x] x/2= 3/4+o(x)

_______________________________________________________

**Special Note for o(x)**

o(x) is a variable dependent on x which satisfies Limit x->0 {o(x)/x}=0.

Hence o(x) is always much smaller than x, for example x²,x³ or sin²x

o(x)+o(x)=o(x)

o(x) o(x)= atleast o(x) or exactly o(x²)

o(x) anyC =o(x)

o(x) x/x =o(x)

_______________________________________________________

Let's now have a look at the whole limit of 3/4+o(x) - it approaches 3/4 when x gets smaller and smaller, and the limit should be 3/4

when n=1+m, 3/4

when n<1+m, no sense or infinity.

when n>1+m, 0.