Answer
$$\lim_{x\to-2^+}\frac{x^2-1}{2x+4}=\infty$$
$$\lim_{x\to-2^-}\frac{x^2-1}{2x+4}=-\infty$$
$$\lim_{x\to1^+}\frac{x^2-1}{2x+4}=0$$
$$\lim_{x\to0^-}\frac{x^2-1}{2x+4}=-\frac{1}{4}$$
Work Step by Step
$$f(x)=\frac{x^2-1}{2x+4}=\frac{(x-1)(x+1)}{2(x+2)}$$
(a) As $x\to-2^+$:
$(x+2)\to0^+\gt0$
$(x+1)\to-1^+\lt0$
$(x-1)\to-3^+\lt0$
So $\frac{(x-1)(x+1)}{2(x+2)}\gt0$ as a result.
That means $\frac{(x-1)(x+1)}{2(x+2)}$ will approach $\infty$. In other words, $$\lim_{x\to-2^+}\frac{x^2-1}{2x+4}=\infty$$
(b) As $x\to-2^-$:
$(x+2)\to0^-\lt0$
$(x+1)\to-1^-\lt0$
$(x-1)\to-3^-\lt0$
So $\frac{(x-1)(x+1)}{2(x+2)}\lt0$ as a result.
That means $\frac{(x-1)(x+1)}{2(x+2)}$ will approach $-\infty$. In other words, $$\lim_{x\to-2^-}\frac{x^2-1}{2x+4}=-\infty$$
(c) $$\lim_{x\to1^+}f(x)=\lim_{x\to1^+}\frac{x^2-1}{2x+4}$$ $$\lim_{x\to1^+}f(x)=\frac{1^2-1}{2\times1+4}=\frac{0}{6}=0$$
(d) $$\lim_{x\to0^-}f(x)=\lim_{x\to0^-}\frac{x^2-1}{2x+4}$$ $$\lim_{x\to0^-}f(x)=\frac{0^2-1}{2\times0+4}=-\frac{1}{4}$$
