Find limit x tends to 0 for { x.log(1+2tanx) }/(1-cosx)

Question

Arun · Answer

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

Aditya Gupta · Answer

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 :)

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Find limit x tends to 0 for { x.log(1+2tanx) }/(1-cosx)

Find limit x tends to 0 for { x.log(1+2tanx) }/(1-cosx)

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