l'hospital's law problem?

Question

How to solve
lim (sinx)^x
x->0+
I know that the limit is 1 but I don't know how to come to it

Yes, but I have to use l'hospital's law. Can you tell me how to do this

Answer 1

y = (sinx)^x is a beautiful function:

http://s210.photobucket.com/albums/bb64/oregfiu/?action=view¤t=sinxx.jpg

Actually we don’t need Mr. l'Hôpital’s help.

We may use formal (“delta-epsilon”) definition of limit.

Since
(sinε)^ε ≤ 1^ε
for any small ε

δ = 1^ε – 1
will do the work.

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lim(sinx)^x
= lim exp(ln((sinx)^x))
= lim exp(x(ln(sinx)))
= exp(lim (x (ln(sinx)))
= exp(lim(ln(sinx))/(1/x))
(using l'Hôpital’s rule)
= exp(lim((1/sinx)*(cosx))/(-1/x²))
= exp((lim x/sinx)*(lim(–x cos x)))
= exp (1*0)
= exp(0)
= 1

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