How to find the Limit of the following Term as x approaches 1?

#(2/(x+1))^(x/(x-1))#
I began like this: #e^(x*ln(2/(x+1))/(x-1))#
Is my beginning right? And how should i go on with LHospital (if Im right)?

Thx for the help

1 Answer
VNVDVI
Apr 9, 2018

#e#

Explanation:

As you know, plugging in #1# right away entails #1^oo,# an indeterminate form, so we will use l'Hospital's Rule. I personally prefer to work it the following way when it comes to these functions, as the logarithm function has some very nice properties:

#y=(2/(x+1))^(x/(x-1))#

#lny=ln((2/(x+1))^(x/(x-1)))#

#lny=(xln(2/(x+1)))/(x-1)#, as #ln(a^b)=blna#

Since our original limit resulted in an indeterminate form, so will this one, as nothing has truly been changed. So, we want #lim_(x->1)lny# and apply l'Hospital's Rule as follows:

#lim_(x->1)(xln(2/(x+1)))/(x-1)=lim_(x->1)((2x)/(x+1)+ln(2/(x+1)))=1+ln(2/2)=1#

So, #lim_(x->0)lny=1#.

We want #lim_(x->1)y#.

From the fact that #y=e^lny#, we can use the limit we already found.

#lim_(x->1)y=lim_(x->1)e^lny=e^(lim_(x->1)lny)=e^1=e#

Related questions
Impact of this question
253 views around the world
You can reuse this answer
Creative Commons License