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

Question

1450 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-05T10:21:04

#= 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