MuPAD doesn't simplify the expression, and thus can't take the limit, because you haven't provided the appropriate assumptions. It's not true that an expression such as (sin(x)^n)^(1/n)
simplifies to sin(x)
if n
and x
are positive. You need to fully restrict the range of x
as it the argument of a periodic function:
x = sym('x','positive');
n = sym('n','real');
assumeAlso(x<=sym(pi));
sn = 8^n * n^2 * (sin(x))^(3*n);
sn2 = simplify(sn^(1/n))
limit(sn2, n, Inf)
which returns the answer you were expecting
ans =
8*sin(x)^3
If you have access to Mathematica, this sort of thing can be accomplished very easily:
Limit[(8^n n^2 Sin[x]^(3 n))^(1/n), n -> \[Infinity], Assumptions -> {n \[Element] Reals, x >= 0, x <= \[Pi]}]