limit

Sol
lim(x->0)_(1－Cosx)/x^2 0/0 type
=lim(x->0)_Sinx/(2x) 　 0/0 type
=lim(x->0)_Cosx/2
=1/2

For trigo limit, the most important thing is that x->0, sinx / x -> 1.

For (1-cosx)/x^2, we can convert cos into sin by the identity sin^2 x + cos^2 x = 1

(1-cosx)/x^2=(1-cox^2 x) / [x^2(1+cos x)]=sin^2 x/ [x^2 (1+cosx)]
factor the sinx / x, [sinx / x]^2 / (1+cosx).
The denominator is no longer zero, sub x = 0,

lim (x->0) [sinx / x]^2 = 1^2 = 1
1/(1+cosx)=1/(1+1)=1/2

lim(x->0) (1-cosx)/x^2
=1/2

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full solution [without explanation]
lim(x->0) (1-cosx)/x^2
=lim(x->0) sin^2 x/ [x^2 (1+cosx)]
=lim(x->0) (sinx/x)^2 / (1+cosx)
=1^2 / 2
=1/2

There are two ways of solving the question.

Method 1: L' Hospital Rule

lim x->0 (1 - cosx) / x^2
= lim x->0 sinx / 2x
= lim x->0 cosx / 2
= cos0 / 2
= 1/2

Please be aware that
lim x->0 (sinx / x) = lim x->0 (x / sin x) = 1

Prove:
When x -> 0,
sinx -> 0 and x -> 0
sinx/x is undefined when x=0
By L' Hospital Rule, we differentiate both the numerator (x) and the denominator (sinx) with respect to x, in order to obtain cosx / 1. When x->0, the result is cos0 = 1


Method 2

lim x->0 (1 - cosx) / x^2
= lim x->0 2(sin x/2)^2 / x^2
= lim x->0 2 [sin(x/2) / x]^2
= 2 lim x->0 [1/2 sin(x/2) / (x/2)]^2
= 2 lim x->0 {1/4 [sin(x/2) / (x/2)]^2}
= 2 * 1/4 lim x->0 [sin(x/2) / (x/2)]^2
= 1/2 lim x->0 [sin(x/2) / (x/2)]^2
= 1/2 {[lim x->0 sin(x/2) / (x/2)]}^2
= 1/2 * (1)^2
= 1/2 * 1
= 1/2

Notice that
(sinx)^2 = 1/2 (1 - cos2x)
(cosx)^2 = 1/2 (1 + cos2x)