Find the solution of lim(x->0) csc^2 (2x) -(1/{4x^2}) without L'hopital?

Setting u=2x gives cosec²(2x)−1/(4x²) = 1/sin²u–1/u² = (u+sinu)(u−sinu) / u²sin²u

= (1+sinu/u) * (u²/sin²u) * (u−sinu)/u³ … (i)

Since lim sinu/u = lim u/sinu = 1 we just need to find lim (u−sinu)/u³

Now sinu = 3sin(u/3)−4sin³(u/3)

So (u−sinu)/u³ = (u−3sin(u/3)+4sin³(u/3)) / u³

Setting v=u/3 in this gives (1/9)(v−sinv)/v³ + (4/27)(sinv/v)³

∴ lim (u→0) (u−sinu)/u³ = (1/9) lim (v→0) (v−sinv)/v³ + (4/27) lim (v→0) (sinv/v)³

Setting lim (z→0) (z−sinz)/z³ = L gives L = L/9 + 4/27 → L=⅙

Hence limit of (i) = 2*⅙ = ⅓

lim(x->0) csc^2 (2x) -(1/{4x^2})

= lim(x->0)[(1/sin²2x) -(1/{(2x)²]

= lim(x->0)[{(2x)² -sin²(2x)}/{(sin²2x)*{(2x)²}]

= lim(x->0)[{((2x)+sin(2x))*(2x)-sin(2x))}/{(sin²2x)*{(2x)²}]

dividing each factor of N by sin 2x and D by sin² 2x we get

= lim(x->0)[{((2x)/sin 2x)} +1}/2x]*[{((2x)/sin 2x)} -1}/2x]

= ∞ because first factor is infinite and second has 0/0 form which can have finite limit, it cannot be zero.