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

1 Answer
Feb 26, 2017

See below.

Explanation:

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])