Question

how do you find the limit of x^sinx as x approaches 0 from the right?

Answer 1

z = lim (sin x)^x
. . .x → 0


ln z = lim x ln (sin x)
. . .x → 0



ln z = lim ln (sin x)/(1/x)
. . .x → 0


using L'Hospital's rule,


ln z = - lim (cos x)(1/sin x)/(1/x²)
. . .x → 0


ln z = - lim x (cos x)(x/sin x)
. . .x → 0



ln z = (0) (1)(1) = 0

z = e^(0)

z = 1

Answer 2

it's \(x^{\sin{x}}\) not \(\sin^xx\)