Re: jnu 2011

Posted by seema on May 15, 2012; 7:51pm
URL: http://discussion-forum.276.s1.nabble.com/jnu-2011-tp7556974p7560578.html

@aastha
 Q 57
given f'(x)= 3x  => f(x) = (3/2)x^2
                          g(x) = (2/3)x^3

solving lim(x -> 1) , u will get 0/0 form. use l'hospital's rule

lim (x ->1) (3x+2x^2)/1 = 5 (option a)

:)