Answer
a) The limit = $\infty$.
b) The limit does not exist.
Work Step by Step
$$f(x)=\frac{x^2+x}{x^5+2x^4+x^3}$$
a) $$\lim_{x\to0}\frac{x^2+x}{x^5+2x^4+x^3}=\lim_{x\to0}\frac{x(x+1)}{x^3(x^2+2x+1)}=\lim_{x\to0}\frac{x+1}{x^2(x+1)^2}$$
$$\lim_{x\to0}\frac{x^2+x}{x^5+2x^4+x^3}=\lim_{x\to0}\frac{1}{x^2(x+1)}$$
As $x\to0^-$ or $x\to0^+$, we see that $x^2(x+1)$ still approaches $0$ from the right, so $x^2(x+1)$. Therefore, function $\frac{1}{x^2(x+1)}$ will approach $\infty$.
$$\lim_{x\to0}\frac{x^2+x}{x^5+2x^4+x^3}=\infty$$
b) $$\lim_{x\to-1}\frac{x^2+x}{x^5+2x^4+x^3}$$
Using the operations in a), we have:
$$\lim_{x\to-1}\frac{x^2+x}{x^5+2x^4+x^3}=\lim_{x\to-1}\frac{1}{x^2(x+1)}$$
As $x\to-1^+$, we have $x^2(x+1)$ approaches $0$ from the right, so $x^2(x+1)\gt0$. Therefore, function $\frac{1}{x^2(x+1)}$ will approach $\infty$.
As $x\to-1^-$, we have $x^2(x+1)$ approaches $0$ from the left, so $x^2(x+1)\lt0$. Therefore, function $\frac{1}{x^2(x+1)}$ will approach $-\infty$.
Since $\frac{1}{x^2(x+1)}$ does not approach the same value as $x\to-1$, $\lim_{x\to-1}\frac{1}{x^2(x+1)}$ does not exist.
Therefore, $\lim_{x\to-1}\frac{x^2+x}{x^5+2x^4+x^3}$ also does not exist.
