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Does #a_n={(3/n)^(1/n)} #converge? If so what is the limit?

1 Answer

Mar 27, 2016

It converges to #1#.

Explanation:

#f(x) = (3/x)^(1/x) = e^(ln(3/x)/x)#

#lim_(xrarroo)(ln(3/x)/x)# has initial form #(-oo)/oo# which is indeterminate, so use l'Hospital's rule.

#lim_(xrarroo) (x/3(-3/x^2))/1 = lim_(xrarr0)-1/x = 0#

Since the exponent goes to #0#, we have

#lim_(xrarroo)(ln(3/x)/x) = e^0 = 1#

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