How do you find the limit of #x(a^(1/x)-1)# as #x->oo#?
2 Answers
Jun 1, 2017
Explanation:
Jun 1, 2017
Explanation:
Rewrite the limit as
#lim_(x->oo)(a^(1/x)-1)/(1/x)#
So that it produces the indeterminate form
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