Verified answer

To solve lim(x->0+)((sin(x))^(sin(x))), we need to look at what ln((sin(x))^(sin(x)))

does as x -> 0+.

So if lim(x->0+)(ln(sin(x))^(sin(x)))) = L, then lim(x->0+)((sin(x))^(sin(x))) = e^L.

(To see this, note that in general, if ln(a) = L then a = e^L for a > 0. It's the

same idea here but with limits involved.)

Now lim(x->0+)(ln(sin(x))^(sin(x)))) =

lim(x->0+)(sin(x) * ln(sin(x))) =

lim(x->0+)(ln(sin(x) / csc(x)), which is in the indeterminate form -infinity/infinity.

Thus we can use L'Hopital's rule:

lim(x->0+)((1/sin(x))*(-cos(x)) / (csc(x)*cot(x))) =

lim(x->0+)(-sin(x)) = 0. So this is the value for L, and thus

lim(x->0+)((sin(x))^(sin(x))) = e^0 = 1.

y=sinx^sinx

lny=sinxlnsinx

let x approach 0.

we get 0*ln0 as a limit and get 0

So lny=0 and y=1 as a limit.

Enanitos Verdes