anonymous
  • anonymous
lim x->inf (-1)^x = ? lim x->inf ((-1)^x)*x = ? lim x->inf x = ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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TuringTest
  • TuringTest
start with the easiest\[\lim_{x\to\infty}x=\]what do you think?
anonymous
  • anonymous
infinity
TuringTest
  • TuringTest
of course!

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experimentX
  • experimentX
the rest are undefined ..
anonymous
  • anonymous
for the first on maple it comes -1-I .. 1+I
TuringTest
  • TuringTest
@experimentX what you just did is called "sniping" - when I am trying to provide an understanding and you burst in with just the answer please do not do that, it lowers the value of the site
TuringTest
  • TuringTest
@woundedtiger40 next most difficult (in my opinion)\[\lim_{x\to\infty}(-1)^x\]now what are some of the terms of this sequence ? what f(x)=(-1)^x what is f(0), f(1), etc...?
anonymous
  • anonymous
@TuringTest please go on
anonymous
  • anonymous
1, -1,1,-1,1..... it's an alternating series
TuringTest
  • TuringTest
so does it ever approach one particular value?
anonymous
  • anonymous
no OK so the limit D.N.E.
TuringTest
  • TuringTest
exactly :D
anonymous
  • anonymous
hence it's product will not exist
anonymous
  • anonymous
am I right
anonymous
  • anonymous
@TuringTest you rock man
anonymous
  • anonymous
thankssssssss
TuringTest
  • TuringTest
yes, because the (-1)^n part is making each point jump from one side of the x-axis to the other, AND f(x)=x does not converge to zero the limit of their product DNE -You Rock WT!
anonymous
  • anonymous
@TuringTest I mean you are good in teaching
anonymous
  • anonymous
explaining things...
TuringTest
  • TuringTest
Thank you, I try :D -\(if\) f(x) \(did\) converge to zero though, then the limit to infty may exist for example\[\lim_{x\to\infty}{(-1)^x\over x}=0\]because even though the whole thing oscillates, the oscillation gets smaller and smaller, eventually just being a straight line (i.e. y=0) http://www.wolframalpha.com/input/?i=plot+%28-1%29%5Ex%2Fx

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