2. \( \lim _{x \rightarrow-2} \frac{8+x^{3}}{4-x^{2}}, \mathrm{R}: 3 \)

To evaluate the limit, first substitute \( x = -2 \) directly into the expression: \[ \lim _{x \rightarrow -2} \frac{8 + x^3}{4 - x^2} = \frac{8 + (-2)^3}{4 - (-2)^2} = \frac{8 - 8}{4 - 4} = \frac{0}{0} \] This is an indeterminate form, so we need to factor the expression. The numerator \( 8 + x^3 \) can be factored as \( (x + 2)(x^2 - 2x + 4) \) using the sum of cubes formula. The denominator \( 4 - x^2 \) can be factored as \( (2 - x)(2 + x) \). Now, rewriting the limit gives us: \[ \lim _{x \rightarrow -2} \frac{(x + 2)(x^2 - 2x + 4)}{(2 - x)(2 + x)} \] Cancel out \( (x + 2) \): \[ \lim _{x \rightarrow -2} \frac{x^2 - 2x + 4}{-(x - 2)} \] Now substitute \( x = -2 \) again: \[ = \frac{(-2)^2 - 2(-2) + 4}{-(-2 - 2)} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3 \] So the limit is 3, confirming your result. To recap: the key here was recognizing this as an indeterminate limit and employing factoring to solve it effectively. It's essential to keep these strategies at your disposal, especially in calculus!