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

Praveen Kumar Das , 5 Years ago
Grade 12th pass
anser 2 Answers
Arun

Last Activity: 5 Years ago

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

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 :)

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