Question

lim x->inf (-1)^x = ?
lim x->inf ((-1)^x)*x = ?
lim x->inf x = ?

Answer 1

start with the easiest\[\lim_{x\to\infty}x=\]what do you think?

Answer 2

infinity

Answer 3

of course!

Answer 4

the rest are undefined ..

Answer 5

for the first on maple it comes -1-I .. 1+I

Answer 6

@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

Answer 7

@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...?

Answer 8

@TuringTest please go on

Answer 9

1, -1,1,-1,1..... it's an alternating series

Answer 10

so does it ever approach one particular value?

Answer 11

no OK so the limit D.N.E.

Answer 12

exactly :D

Answer 13

hence it's product will not exist

Answer 14

am I right

Answer 15

@TuringTest you rock man

Answer 16

thankssssssss

Answer 17

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!

Answer 18

@TuringTest I mean you are good in teaching

Answer 19

explaining things...

Answer 20

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