Lim(x-0)[1/x]^sinx

Question

Lim(x->0)[1/x]^sinx

anonymous · Answer

that form is $$\infty$$^0, so u have to use exp function

anonymous · Answer

The answer given is 1

anonymous · Answer

Why don't you do a series expansion in x and then send the function to zero?

anonymous · Answer

Actually, I should have worked it out before opening my mouth.

anonymous · Answer

exp[lim(x->0) sin x . 1/x] 
exp[lim(x->0) (ln x)/sin x]
and then do l'hospital

anonymous · Answer

Actually, what I said will work, but you'd have to prove other limits first.

anonymous · Answer

i do thinking about the concept before typing

anonymous · Answer

What are you thinking, iamignorant?

anonymous · Answer

Yes, I have got the idea that suzi gave. With that idea I am coming at e^(0x1)

anonymous · Answer

So that is the answer that I needed. Actually I did everything he did except applying lHospital

anonymous · Answer

ok

anonymous · Answer

Thank you all

anonymous · Answer

np

anonymous · Answer

is there a rank higher than HERO? =D like a grandmaster or something

anonymous · Answer

Who knows?  This site's operations are a mystery half the time.

anonymous · Answer

i think GOD

anonymous · Answer

=D LOL

anonymous · Answer

It's dead today...

anonymous · Answer

Incidentally, I set the RHS equal to y then took the log of both sides, put the result into an indeterminate form, used L'Hopital's rule and took the limit to zero after I suggested expanding the function.  Just sayin'...

anonymous · Answer

What's going on BMF?