Does #lim_(xrarroo) n^x = oo#?

1 Answer
Jim H · Alan P. · MysticSmile
Oct 1, 2017

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#

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