What is the limit of #x^(1/x) # as x approaches infinity?

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Answer 1 · 2016-04-06T11:52:31

#lim_(xrarroo)x^(1/x) = 1#

#lim_(xrarroo)x^(1/x) = lim_(xrarroo)e^ln(x^(1/x))#

# = lim_(xrarroo)e^(1/xlnx)#

# = e^(lim_(xrarroo)(1/xlnx))#.

And

#lim_(xrarroo)(1/xlnx) = lim_(xrarroo)(lnx/x)# which has indeterminate form #oo/oo#.

Apply l'Hospital's Rule:

#lim_(xrarroo)(lnx/x) = lim_(xrarroo)((1/x)/1) = 0#

Since the exponent goes to #0#, we have

#lim_(xrarroo)x^(1/x) = lim_(xrarroo)e^(1/xlnx)#

# = e^0 = 1#