You may evaluate the limit using the half of angle formula such that:

`1 - cos x = 2 sin^2(x/2)`

You need to substitute `2 sin^2(x/2)`  for `1 - cos x`  in equation under limit such that:

`lim_(x-gt0) (2 sin^2(x/2))/(x^3+x^2)`

You need to use special limit `lim_(x-gt0) sin x/x = 1` , hence, you may form special limit such that:

`lim_(x-gt0) (2 sin^2(x/2))/(x^3+x^2) = 2lim_(x-gt0) ((sin^2(x/2))(x^2/4))/((x^2/4)(x^3+x^2))`

`lim_(x-gt0) (2 sin^2(x/2))/(x^3+x^2) = (2/4) lim_(x-gt0) (sin^2(x/2))/(x^2/4)*lim_(x-gt0) x^2/((x^3+x^2))`

`lim_(x-gt0) (2 sin^2(x/2))/(x^3+x^2) = (2/4)*1*lim_(x-gt0) x^2/((x^3+x^2))`

Notice that the limit `lim_(x-gt0) x^2/((x^3+x^2)) = 0/0`  is indetermination, hence you may use l'Hospital's theorem such that:

`lim_(x-gt0) (x^2)/((x^3+x^2)) = lim_(x-gt0) ((x^2)')/((x^3+x^2)')`

`lim_(x-gt0) ((x^2)')/((x^3+x^2)') = lim_(x-gt0) (2x)/(3x^2+2x) `

`lim_(x-gt0) (2x)/(3x^2+2x) = lim_(x-gt0) ((2x)')/((3x^2+2x)')`

`lim_(x-gt0) ((2x)')/((3x^2+2x)') = lim_(x-gt0) 2/(6x+2)`

You need to substitute 0 for x in equation under limit such that:

`lim_(x-gt0) 2/(6x+2) = 2/(0+2) = 2/2 = 1`

`lim_(x-gt0) (2 sin^2(x/2))/(x^3+x^2) = (2/4)*1*1 = 1/2`

Hence, evaluating the limit to the function yields `lim_(x-gt0) (1- cos x)/(x^3+x^2)`  yields  `lim_(x-gt0) (1- cos x)/(x^3+x^2)= 1/2.`

The limit `lim_(x->0)(1 - cos x)/(x^3 + x^2)` has to be determined.

Substituting x = 0 in `lim_(x->0)(1 - cos x)/(x^3 + x^2)` gives the form `0/0` which is indeterminate. This allows the use of the l'Hopital's Theorem and the numerator and denominator can be substituted with their derivatives.

=> ` lim_(x->0)(sin x)/(3x^2 + 2x)`

Substituting x = 0 gives an indeterminate form `0/0` . Again replace the denominator and numerator by their derivatives

=> ` lim_(x->0)(cos x)/(6x + 2)`

Substituting x = 0 gives `1/2`

The value of `lim_(x->0)(1 - cos x)/(x^3 + x^2)` = `1/2`