How do you find the limit of #(1+(7/x)+(3/x^2))^x# as x approaches infinity?

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Answer 1 · 2017-02-26T20:52:36

See below.

Solving #1+7/x+3/x^2=3(1/x+r_1)(1/x+r_2)# we have

#(1+7/x+3/x^2)=3 r_1 r_2(r_1/x+1)(r_2/x+1)# and also

#(1+7/x+3/x^2)^x=(3 r_1 r_2)^x(r_1/x+1)^x(r_2/x+1)^x#

but #3 r_1 r_2 = 1# so

#lim_(x->oo)(1+7/x+3/x^2)^x = lim_(y_1->oo)(1/y_1+1)^(y_1/r1)lim_(y_2->oo)(1/y_2+1)^(y_2/r_2)#

giving

#lim_(x->oo)(1+7/x+3/x^2)^x=e^(1/r_1) e^(1/r_2)= e^(1/r_1+1/r_2)#

with #r_1 = -1/6 (7 + sqrt[37])# and #r_2 = -1/6 (7 - sqrt[37])#