What is #1^oo# ?
1 Answer
Explanation:
We are reasonably safe with this one, though if you tweak it in any way it can get messy.
What do we mean by
Perhaps you think of:
#1^n = overbrace(1 * 1 * ... * 1)^(n " times")#
So:
#1^oo = overbrace(1 * 1 * ... * 1)^(oo " times")#
That's not strictly accurate.
#1^oo = lim_(n->oo) 1^n = lim_(n->oo) 1 = 1#
Footnote
To see some of the dangers in limit calculations, consider:
#lim_(n->oo) (1+1/n)^n#
We might naively put:
#lim_(n->oo) (1+1/n)^n = (1+1/oo)^oo = (1+0)^oo = 1^oo = 1" "# WRONG
Instead we need can find the true value by looking at the actual value for finite values of
Using the binomial theorem we actually find:
#lim_(n->oo) (1+1/n)^n = sum_(k=0)^oo 1/(k!) = e#