How do you find the limit of #x(a^(1/x)-1)# as #x->oo#?

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Answer 1 · 2017-06-01T19:10:58

# log_e a#

#x(a^(1/x)-1)= (a^(0+1/x)-a^0)/(1/x)# now calling #1/x = h# we have

#lim_(x->oo)x(a^(1/x)-1)=lim_(h->0)(a^(0+h)-a^0)/h = a^0log_e a=log_e a#

Answer 2 · 2017-06-01T19:13:13

#ln(a)#

Rewrite the limit as

#lim_(x->oo)(a^(1/x)-1)/(1/x)#

So that it produces the indeterminate form #0/0#.

Now use L'hopital's rule (and chain rule as a result):

#lim_(x->oo)(a^(1/x)-1)/(1/x)=lim_(x->oo)(a^(1/x) * ln(a) * (-1)/x^2)/((-1)/x^2)#

#=lim_(x->oo)a^(1/x)ln(a) = a^0ln(a) = ln(a)#

Final Answer