Calculate the limit (1-cosx)/x^2

, x->0

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The limit `lim_(x->0) (1-cosx)/x^2` has to be determined.

If we substitute x = 0, the result `(1-cosx)/x^2` is the indeterminate form `0/0` . In this case it is possible to use l'Hospital's rule and replace the numerator and denominator with their derivatives.

`(1 - cos x)' = sin x`

`(x^2)' = 2x`

The limit is now:

`lim_(x->0) sin x/(2x)`

If the substitution x = 0 is made here, the result is again the indeterminate form `0/0` . Using l'Hospital's rule, replace the numerator and denominator with their derivatives. This gives:

`lim_(x->0) cos x/2`

Substituting x = 0 gives the result 1/2.

The limit `lim_(x->0) (1-cosx)/x^2 = 1/2` .

We'll substitute the numerator:

1 - cos x = 2 [sin(x/2)]^2

Lim (1-cosx)/x^2 = Lim {2 [sin(x/2)]^2}/x^2

Lim {2 [sin(x/2)]^2}/x^2 = Lim [sin(x/2)]^2/(x^2/2)

Lim [sin(x/2)]^2/(x^2/2) = lim[sin(x/2)/(x/2)]*lim[sin(x/2)/2*(x/2)]

Since the elementary limit is:

lim [sin f(x)]/f(x) = 1, if f(x) -> 0, we'll get:

**Lim (1-cosx)/x^2 = 1/2**

To find the Lt x-> 0 (1-cosx)/x^2.

Cosx = 1-2 (sinx/2)^2.

1-cosx = 1-(1-(2sinx/2)^2.

1-cosx = 2(sinx/2)^2.

Therefore Lt x--> 0 (1-cosx)/x^2 = Lt x-->0 2(sinx/2)^2/x^2.

Lt x--> 0 (1-cosx)/x^2 = 2Lt y-->0 (siny)^2/(2y)^2 , where y = x/2.

Lt x--> 0 (1-cosx)/x^2 = Lt y-->0 (2/4)* [(siny)/y ]^2.

Lt x--> 0 (1-cosx)/x^2 = 2/4 , as Lt y-->0 (siny)/y = 1.

Therefore Lt x--> 0 (1-cosx)/x^2 = 1/2.

lim x→0 (1-cosx)/x²

= lim x→0 [(1-cosx)(1+cosx)]/[x²(1+cosx)]

= {lim x→0 sinx/x}². {limx→0 1/(1+cosx)}

= 1² . 1/(1+1)

= 1/2

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