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Find limit x tends to 0 for { x.log(1+2tanx) }/(1-cosx)
Dear student x² log ( 1 + 2 tanx) *2 tanx/x* 2 tanx(1 - cosx) Now put limit 1 * 2 * 2 = 4 Hope it helps Thanks
write it as[x/(1 – cosx)]*[log(1+2tanx)/2tanx]*2tanxso, 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= 4kindly approve :)
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