Re: Limits doubt

Posted by mrittik on Jun 21, 2014; 6:57am
URL: http://discussion-forum.276.s1.nabble.com/Limits-doubt-tp7593191p7593255.html

its a form of 0/0...so u can easily use the l'hospital rule here...now the limiting fn look like:

a^xlna-ax^(a-1)/(1+lnx)x^x=a^alna-a^a/(1+lna)a^a=lna-1/1+lna=-a=> 1-lna/1+lna=a

1-lna=a+alna=>1-a=(1+a)lna=> 2/(1+a)=lna...it comes like this..plz chk once