what is limit of #x*log((1+e^x)/((-1)+e^x)))# as #x# approaches infinity ??

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Mar 13, 2018

Answer:

#0#

Explanation:

#(e^x + 1)/(e^x - 1) = (1 + e^-x)/(1 - e^-x)#
#= (1 + e^-x)^2/((1 - e^-x)(1 + e^-x))#
#= (1 + e^-x)^2/(1 - e^(-2 x))#
#= (1 + 2 e^-x + e^(-2 x))/(1 - e^(- 2 x))#
#~~ 1 + 2 e^-x#

#log(1+x) = x - x^2/2 + x^3/3 - ..." (Taylor series of ln(1+x))"#

#=> log((e^x + 1)/(e^x - 1)) ~~ 2 e^-x - 2 e^(-2 x) + ...#

#=> x log( ... ) ~~ 2 x e^-x -> 0" for "x -> oo#

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Mar 13, 2018

Answer:

#lim_(x->oo)(x*log[(1+e^x)/(-1+e^x)])=0#

Explanation:

Let's think about it this way:

As #x# gets really, really, large, #e^x# will be so large that numbers like #1# and #-1# will barely matter.

We can, therefore, ignore it.

We are left with #e^x/e^x# Since we are dividing same numbers, we conclude that #lim_(x->oo)(1+e^x)/(-1+e^x)=1#

We have:

#x*log(1)# Simplify

#=>x*0#

#=>0# The answer is #0#.

Just note that this method may not work as you go to harder versions of limit.

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