Year 12s: what will be the first way you research your future options? >>

(-1)^∞

well actually i am concernded with, and confused by;

(-1)^x, as as x tends to ∞

or any

(-y)^x, as as x tends to ∞

:confused:

Thanks

Reply 1

nuodai

What set is

x

in? Is it integers or real numbers (or even complex numbers)?

Either way it makes little difference though, to be honest...

(-1)^x

= -1 for odd x and 1 for even x, and is complex/undefined for

x = \frac{1}{2n}

, and since

\infty

is neither odd nor even, it has no limit (unless there's something I'm missing). Same goes for

(-y)^x

Reply 2

Chewwy

Yarikh

well actually i am concernded with, and confused by;

(-1)^x, as as x tends to ∞

or any

(-y)^x, as as x tends to ∞

:confused:

Thanks

seriously, do you think a limit exists?

Reply 3

Conor Tickner

It is undefined. Whether (-1)^x is -1 or 1 depends on whether x is odd or even. ∞ is neither so it's undefined. It doesn't tend to anything as it's an oscilating sequence.

Reply 4

generalebriety

Chewwy

seriously, do you think a limit exists?

This.

I don't mean to be rude, but if it looks like an answer can't exist or your question just doesn't make any sense, I don't see why you're posting it. Clearly (-1)^n keeps flipping signs as n increases and will keep doing so no matter how big n is.