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- anonymous

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

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- anonymous

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

- schrodinger

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- anonymous

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

- anonymous

The answer given is 1

- anonymous

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

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- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

i do thinking about the concept before typing

- anonymous

What are you thinking, iamignorant?

- anonymous

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

- anonymous

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

- anonymous

ok

- anonymous

Thank you all

- anonymous

np

- anonymous

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

- anonymous

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

- anonymous

i think GOD

- anonymous

=D LOL

- anonymous

It's dead today...

- anonymous

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

What's going on BMF?

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