anonymous
  • anonymous
Lim(x->0)[1/x]^sinx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
that form is \[\infty\]^0, so u have to use exp function
anonymous
  • anonymous
The answer given is 1
anonymous
  • anonymous
Why don't you do a series expansion in x and then send the function to zero?

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anonymous
  • anonymous
Actually, I should have worked it out before opening my mouth.
anonymous
  • anonymous
exp[lim(x->0) sin x . 1/x] exp[lim(x->0) (ln x)/sin x] and then do l'hospital
anonymous
  • anonymous
Actually, what I said will work, but you'd have to prove other limits first.
anonymous
  • anonymous
i do thinking about the concept before typing
anonymous
  • anonymous
What are you thinking, iamignorant?
anonymous
  • anonymous
Yes, I have got the idea that suzi gave. With that idea I am coming at e^(0x1)
anonymous
  • anonymous
So that is the answer that I needed. Actually I did everything he did except applying lHospital
anonymous
  • anonymous
ok
anonymous
  • anonymous
Thank you all
anonymous
  • anonymous
np
anonymous
  • anonymous
is there a rank higher than HERO? =D like a grandmaster or something
anonymous
  • anonymous
Who knows? This site's operations are a mystery half the time.
anonymous
  • anonymous
i think GOD
anonymous
  • anonymous
=D LOL
anonymous
  • anonymous
It's dead today...
anonymous
  • 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
  • anonymous
What's going on BMF?

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