Aditya Gupta
Last Activity: 5 Years ago
write it as
[x/(1 – cosx)]*[log(1+2tanx)/2tanx]*2tanx
so, lim becomes (limit x tends to 0 2xtanx/(1 – cosx))*(limit x tends to 0 log(1+2tanx)/2tanx)
= (limit x tends to 0 2xtanx/(1 – cosx))*(limit y tends to 0 log(1+y)/y).....here we have let y= 2tanx
= (limit x tends to 0 2xtanx/(1 – cosx))*1
= limit x tends to 0 2xsinx/cosx(1 – cosx)
= limit x tends to 0 2xsinx(1+cosx)/cosx(1 – cos^2x)
= limit x tends to 0 2xsinx(1+cosx)/cosxsin^2x
= limit x tends to 0 2x(1+cosx)/cosxsinx
= limit x tends to 0 [2(1+cosx)/cosx] * limit x tends to 0 x/sinx
= 4*1
= 4
kindly approve :)