Does lim_(xrarroo) n^x = oo?

1 Answer

Infinity is not a number. But see below.

Explanation:

For n > 1
color(white)("xxx")lim_(xrarroo)n^x = oo

For 0 < n < 1
color(white)("xxx")lim_(xrarroo)n^x = 0

For n=1
color(white)("xxx")lim_(xrarroo)1^x = lim_(xrarroo) 1 = 1
This is true only if the base really is 1 and not some function that approaches 1 as x increases without bound (as xrarroo)

Is the base is a function that approaches 1 as xrarroo, then the form 1^oo is indeterminate.

That is:

If lim_(xrarroo)g(x) = 1 and lim_(xrarroo)f(x) = oo, then

lim_(xrarroo)(g(x))^f(x) is indeterminate.

For example lim_(xrarroo)(1+1/x)^x = e and lim_(xrarroo)(1+5/x)^x = e^5