6 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.did you try L'Hôpital's?
haa wahi doubtful hora ki kaise hpga
limit ko ln(x)/cos(pi/2^x) isko ln(x)/sin(pi/2(2^(x-1)-1)) Now since x tends to 1 ln(x) tends to (x-1). So this tends to (x-1)/(pi/2(2^x-1)) so you get the answer as 2/pi.ln2
OHHHHHHHH OK