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

Question

Emma Aldridge · Accepted Answer

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])#

Luke Alvarez · Answer

undefined To find the limit of (1+(7/x)+(3/x^2))^x as x approaches infinity, we can use the concept of exponential limits.

First, we rewrite the expression as (1+(7/x)+(3/x^2))^x = ((x/x) + (7/x) + (3/x^2))^x.

Next, we simplify the expression inside the parentheses: (x + 7 + 3/x)^x.

As x approaches infinity, the terms 7/x and 3/x^2 approach zero, leaving us with (x + 0 + 0)^x = x^x.

Finally, we evaluate the limit of x^x as x approaches infinity. This limit is indeterminate, so we can apply L'Hôpital's rule.

Differentiating both the numerator and denominator with respect to x, we get (ln(x) + 1) * x^x / (1/x) = (ln(x) + 1) * x^(x+1).

As x approaches infinity, ln(x) approaches infinity, and x^(x+1) also approaches infinity.

Therefore, the limit of (1+(7/x)+(3/x^2))^x as x approaches infinity is infinity.