If substitute x by -2, the limit is indeterminate.

`lim_(x-gt-2) ((x)^5 + 32)/(x+2) = ((-2)^5 + 32)/(-2+2) = (-32+32)/(-2+2) = 0/0`

Use the formula`a^n + b^n = (a+b)(a^(n-1) - a^(n-2)*b + ... + (-1)^(n-1)*b^(n-1))`

`` `x^5 + 32 = (x+2)(x^4 - 2x^3 + 4x^2 - 8x + 16)`

`lim_(x-gt-2) ((x)^5 + 32)/(x+2) = lim_(x-gt-2) ((x+2)(x^4 - 2x^3 + 4x^2 - 8x + 16))/(x+2)`

`` Reduce by x+2:

`lim_(x-gt-2) (x^4 - 2x^3 + 4x^2 - 8x + 16) = (-2)^4 - 2(-2)^3 + 4(-2)^2 - 8(-2) + 16`

`lim_(x-gt-2) (x^4 - 2x^3 + 4x^2 - 8x + 16) = 16 + 16 + 16 + 16 + 16 = 5*16 = 80`

Since the indetermination is 0/0, you may also make use of the Hospital's rule.

`lim_(x-gt-2) ((x)^5 + 32)/(x+2) = lim_(x-gt-2) ((x^5 + 32)')/((x+2)')`

`` `lim_(x-gt-2) (((x)^5 + 32)')/((x+2)') = lim_(x-gt-2) 5x^4 = 5*(-2)^4 = 5*16 = 80`

**ANSWER: `lim_(x-gt-2) ((x)^5 + 32)/(x+2) = 80` **

The limit `lim_(x->-2)(x^5+32)/(x+2)` has to be determined.

If we substitute x = -2 in `(x^5+32)/(x+2)` the result is `(-2^5 + 32)/(-2+2) = 0/0 ` which is indeterminate. This is also i a form that allows us to use l'Hospital's rule and replace the numerator and denominator by their derivatives.

`(x^5 + 32)' = 5x^4`

`(x + 2)' = 1`

The limit can be written as:

`lim_(x -> -2) (5x^4)/1`

Substituting x = -2 gives 5*16 = 80

The required limit `lim_(x->-2)(x^5+32)/(x+2) = 80`