Find the Limit of tanx - sin x

[Math Tutor]

Find the limit:
[tex]\lim_{x\to0} \frac{\tan x-\sin x}{sin^2x}[/tex]

[adamsmith]

Given lim x->0 (tan x-sin x)/sin²x
lim x->0 [(sinx/cos x)-sin x]/sin² x
lim x->0 [sin x/(cos x sin²x)-sin x /sin²x]
lim x->0[1/(cos x sin x)-1/sin x]
multiply and divide by x in the denominator
lim x->0 [1/0-1/0]
applying limit we get
= ∞-∞
=∞

[Guest]

[tex]\frac{tanx-sinx}{ [(sinx)X(sinx)}=

=\frac{sinx(1/cosx-1)}{(sinx)X(sinx) }

=\frac{1-cosx}{ sinx X cosx}

=\frac{2sin^2(x/2)}{2sin(x/2)cos(x/2)cosx }

=\frac{sinx/2}{ cos(x/2)cosx}[/tex]

so for lim x tends 0 the function =0/1=0

[jackwilson]

Full equation with step by step solving thanks for it.