Lim x tends to infinity 2 ^ n sin (x / 2 ^ n)

Lim x tends to infinity 2 ^ n sin (x / 2 ^ n)

When x tends to infinity, sin (x / 2 ^ n) is equivalent to X / 2 ^ n
So lim2 ^ n sin (x / 2 ^ n) = lim2 ^ n * x / 2 ^ n = x = infinity

Lim n tends to infinity 2 ^ n sin (x / 2 ^ n)
X is a constant not equal to 0

2 ^ n sin (x / 2 ^ n) = [sin (x / 2 ^ n) / (x / 2 ^ n)] * x, when n tends to infinity, X / 2 ^ n tends to infinitesimal, so the whole formula tends to X

LIM (n tends to infinity) n π * sin (n π)

sin(nπ）=0
So the limit is equal to 0
In addition: the approach of n to infinity is the approach of rounding