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How would one find

limx0((1+x)1/xe)1/x

using L'Hospital's rule?

 Nov 15, 2016

Best Answer 

 #4
avatar+118710 
+12

Very impressive Heureka :)

 Nov 15, 2016
 #1
avatar+118710 
0
 Nov 15, 2016
 #3
avatar+26398 
+15

How using L'Hospital's rule?

 

limx0((1+x)1xe)1x

 

 

Formula: ln(limx0f(x))=limx0lnf(x)

 

 

1. We  take the logarithm:

ln(limx0((1+x)1xe)1x)=limx0ln{((1+x)1xe)1x}=limx01xln((1+x)1xe)=limx01x[ln((1+x)1x)ln(e)]|ln(e)=1=limx01x[ln((1+x)1x)1]=limx01x[1xln(1+x)1]=limx01x[1xln(1+x)xx]=limx01x[ln(1+x)xx]=limx0ln(1+x)xx2|L'Hospital's rule=limx0(ln(1+x)x)(x2)=limx011+x12x=limx011x1+x2x=limx0x1+x2x=limx0x2x(1+x)=limx012(1+x)|x=0=12(1+0)=12

 

2. We revert the logarithm:

ln(limx0((1+x)1xe)1x)=12eln(limx0((1+x)1xe)1x)=e12limx0((1+x)1xe)1x=e12=1e12limx0((1+x)1xe)1x=1e

 

 

laugh

 Nov 15, 2016
 #6
avatar+7 
0

This may be a slightly late response, but yes, that's correct. Thank you.

I did manage to determine the solution in the time that I was waiting for the answer, as the assignment was due the next day, but thank you nonetheless!

Tylertronics  Nov 17, 2016
 #4
avatar+118710 
+12
Best Answer

Very impressive Heureka :)

Melody Nov 15, 2016
 #5
avatar+26398 
+10

Thank you Melody.

 

smiley

heureka  Nov 15, 2016

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