Can you evaluate this limit?

Yikes! The answer is nice, but finding it is a bear! L'Hopital will work, of course, but you need to apply it more than half a dozen times. (Dozens of chain rules without making a mistake...)

You might try numerical approximation, but both the numerator and denominator get really small really fast, so it is hard to calculate accurate numbers when "x" is close to zero (every x-value much smaller than one hundredth incorrectly gives "undefined"). A spreadsheet approach is not very convincing.

Sketching a graph is fairly convincing. Just don't zoom in TOO much or you start to see weird things (resulting from round off errors).

My best suggestion for exact results is to use Maclaurin Series. Unfortunately, you need to use at least FOUR terms:

Sin(x) = x - x^3 / 6 + x^5 /120 - x^7 / 5040 + O(x^9)

Tan(x) = x + x^3 / 3 + 2x^5 / 15 + 17x^7 / 315 + O(x^9)

ArcSin(x) = x + x^3 / 6 + 3x^5 /40 + 5x^7 / 112 + O(x^9)

ArcTan(x) = x - x^3 / 3 + x^5 / 5 - x^7 / 7 + O(x^9)

Substitute to find Sin(Tan(x)) and Tan(Sin(x)) and ArcSin(ArcTan(x)) and ArcTan(ArcSin(x)). If you can do that (the x^7 term is the killer and it IS needed), then it is easy to subtract and finally divide to get a linear approximation for:

[sin(tan(x)) - tan(sin(x))] / [arcsin(arctan(x)) - arctan(arcsin(x))]

THEN the limit is trivial, just substitute x=0.

As we are left with 0/0, we should use L'Hopitals rule:

a = sin(tan(x))

b = -tan(sin(x))

c = arcsin(arctan(x))

d = -arctan(arcsin(x))

da = cos(tan(x)) * (sec(x)^2)

db = -sec(sin(x))^2 * (cos(x))

dc = 1 / sqrt(1 - arctan(x)^2) * (1 / (1 + x^2))

db = -1 / (1 + arcsin(x)^2) * 1 / sqrt(1 - x^2))

x = 0

da = cos(0) * 1/1 = 1

db = 0 * 1 = 0

dc = 1 / sqrt(1 - 0)) * (1 / (1 + 0)) = 1 / 1 * 1 / 1 = 1

db = -1 / (1 + 0)) * 1 / 1 = -1

1 / (1 - 1) = 1 / 0

Yeah, this is real fing easy

You're going to have to take the 2nd derivative and see if anything changes. But I've wasted too much time on this already.