For x in (0, π/2), (sin(x))^x = [(sin(x))^(sin(x)]^(x/sin(x))
We know that lim t → 0 t^t = 1. The sine function is continuous and sin(0) = 0. Also, sine is nonzero near 0 for x ≠0. Hence, it folows from a well known theorem about limits of compositions of functions that lim x → 0 [(sin(x))^(sin(x)] = 1. In addtion, we have the classic limit lim x → 0 x/sin(x) = 1. Hence, lim x → 0 (sin(x))^x = 1^1 = 1.
To see that lim t → 0 t^t = 1, observe that t^t = exp(t ln(t)). lim t → 0+ t ln(t) = lim u →,-∞ u exp(u) = lim u → ∞ -u/exp(u) = 0. Therefore, lim t → 0 t^t = exp(0) = 1.
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For x in (0, π/2), (sin(x))^x = [(sin(x))^(sin(x)]^(x/sin(x))
We know that lim t → 0 t^t = 1. The sine function is continuous and sin(0) = 0. Also, sine is nonzero near 0 for x ≠0. Hence, it folows from a well known theorem about limits of compositions of functions that lim x → 0 [(sin(x))^(sin(x)] = 1. In addtion, we have the classic limit lim x → 0 x/sin(x) = 1. Hence, lim x → 0 (sin(x))^x = 1^1 = 1.
To see that lim t → 0 t^t = 1, observe that t^t = exp(t ln(t)). lim t → 0+ t ln(t) = lim u →,-∞ u exp(u) = lim u → ∞ -u/exp(u) = 0. Therefore, lim t → 0 t^t = exp(0) = 1.
-1 < sin(x) < 1 for all x in R
(-1)^x < sin(x)^x < 1^x for all x in R
as x - > 0, (-1)^x - > 1
as x - > 0, 1^x - > 1
Therefore, by the Squeeze Theorem of Limits,
as x - > 0, sin(x)^x - > 1
Note that -1 <= sin(x) <= 1 for all real values of x.
Thus, we have:
(-1)^x <= (sin(x))^x <= 1^x
lim(x->0) (-1)^x = 0
lim(x->0) 1^x = 0
Hence, by squeeze theorem, lim(x->0) (sin(x))^x = 0 as well.