You need to evaluate the given limit, hence, you need to replace -1 for x in equation, such that:

`lim_(x->(-1)) (pi - arccos x)^2/(x + 1) = (pi - arcccos(-1))^2/(-1 + 1)`

`lim_(x->(-1)) (pi - arccos x)^2/(x + 1) = (pi - pi + arccos 1)/0`

`lim_(x->(-1)) (pi - arccos x)^2/(x + 1) = (pi - pi + 0)/0`

`lim_(x->(-1)) (pi - arccos x)^2/(x + 1)= 0/0`

You may use l'Hospital's theorem, such that:

`lim_(x->(-1)) ((pi - arccos x)^2)'/((x + 1)')=im_(x->(-1)) (2(pi - arccos x)*1/sqrt(1 - x^2)/1`

`lim_(x->(-1)) 2(pi - arccos x)/sqrt(1 - x^2) = 0/0`

Using again l'Hospital's theorem, yields:

`2lim_(x->(-1)) (1/sqrt(1 - x^2))/-x/(sqrt(1 - x^2)) `

Reducing duplicate terms yields:

`2lim_(x->(-1)) -1/x = 2*(-1/(-1)) = 2`

**Hence, evaluating the given limit, using l'Hospital's theorem, yields **`lim_(x->(-1)) (pi - arccos x)^2/(x + 1) = 2.`