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that form is \[\infty\]^0, so u have to use exp function
The answer given is 1
Why don't you do a series expansion in x and then send the function to zero?
Actually, I should have worked it out before opening my mouth.
exp[lim(x->0) sin x . 1/x] exp[lim(x->0) (ln x)/sin x] and then do l'hospital
Actually, what I said will work, but you'd have to prove other limits first.
i do thinking about the concept before typing
What are you thinking, iamignorant?
Yes, I have got the idea that suzi gave. With that idea I am coming at e^(0x1)
So that is the answer that I needed. Actually I did everything he did except applying lHospital
Thank you all
is there a rank higher than HERO? =D like a grandmaster or something
Who knows? This site's operations are a mystery half the time.
i think GOD
It's dead today...
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'...
What's going on BMF?