How do you evaluate the limit #(x/(x+2))^x# as x approaches #oo#?

Question

Christopher Agresta · Accepted Answer

#= 1/e^2#

#lim_{x to oo} (x/(x+2))^x#

there is a well known limit #lim_{n to oo) (1 + 1/n)^n = e# ....Bernoulli's compounding formula so we can aim for that maybe

#=lim_{x to oo} 1/((x+2)/x)^(x)#

#=lim_{x to oo} 1/(1+2/x)^(x)#

with sub #2/x = 1/y# , #x = 2y#

#=lim_{x to oo} 1/(1 + 1/y)^(2y)#

using the power law of limits #=( lim_{x to oo} 1/(1 + 1/y)^(y))^2#

using the division law #=(( lim_{x to oo} 1)/(lim_{x to oo}(1 + 1/y)^(y)))^2#

#= 1/e^2#

the law quoted are summarised here

IACOBUS BERNOULLI MATHEMATICUS INCOMPARABILIS

Ezra Adams · Answer

undefined To evaluate the limit of (x/(x+2))^x as x approaches infinity, we can rewrite the expression using the natural logarithm. Taking the natural logarithm of both sides, we get ln((x/(x+2))^x). Using the properties of logarithms, we can simplify this expression to x ln(x/(x+2)).

Next, we can apply the limit properties. As x approaches infinity, ln(x/(x+2)) approaches ln(1) which is 0. Therefore, the limit of x ln(x/(x+2)) as x approaches infinity is infinity multiplied by 0, which is an indeterminate form.

To further evaluate this limit, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (1/(x+2)) / (1 - x/(x+2)). Simplifying this expression, we have (1/(x+2)) / (2/(x+2)).

Simplifying further, we get 1/2 as x approaches infinity. Therefore, the limit of (x/(x+2))^x as x approaches infinity is 1/2.