How to evaluate a limit with subtractions $\lim_{x \rightarrow -1}(\frac{3}{x^3+1}-\frac{1}{x+1})$? I'm having trouble thinking of a way to solve this.
$$\lim_{x \rightarrow -1}\left(\frac{3}{x^3+1}-\frac{1}{x+1}\right)$$

Because$$x^3+1=(x+1)(x^2-x+1)$$we get
$$\frac{3}{x^3+1}-\frac{1}{x+1}=\frac{3}{(x+1)(x^2-x+1)}-\frac{1}{x+1}$$
$$=\frac{3-(x^2-x+1)}{(x+1)(x^2-x+1)}=\frac{x-x^2+2}{(x+1)(x^2-x+1)}=$$
$$=\frac{x+1-(x^2-1)}{(x+1)(x^2-x+1)}=\frac{(x+1)(2-x)}{(x+1)(x^2-x+1)}=\frac{2-x}{x^2-x+1}$$
