1(1-x2cot2x)/x2
L'Hospital cant be applied becuz numerator and denominator are not tending to 0
5 Answers
1357(1-x2cot2x)/x2 Applying L'Hospital
=-cot2x+xcotxcosec2x
=(cosx/sinx)(x/sin2x - cosx/sinx)
=(cosx/sin3x)(x-sinxcosx)=(cosx/sin3x)(x-sin2x/2)
Applying limits wherever possible and then applying L Hospital
(1-cos2x)/3sin2x=2sin2x/3sin2x=2/3
1
1@above
arey simplify it na in terms of cos and sin..u will see that its a 0/0 form
1u can also use series expansion to do it,
[sin2x-x2cos2x]/x2sin2x
we have[\left(x-\frac{x^{3}}{3!} \right)^{2}-x^{2}\left(1-\frac{x^{2}}{2!} \right)^{2}}]/[x^{2}\left(x-x^{3}/3! \right)^{2}]
applying binomial theoram,
= [x^{2}(1-2x^{2}/3!)-x^{2}(1-2x^{2}/2!)]/[x^{4}(1-2x^{2}/3!)]
simplifying this we have 2/3 as the final answer
cheers!!
1357[(1-(x/tanx)2]/x2
I considered (x/tanx) tending to 1 and hence applied L'Hospital