## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to-2^+}\frac{x^2-1}{2x+4}=\infty$$ $$\lim_{x\to-2^-}\frac{x^2-1}{2x+4}=-\infty$$ $$\lim_{x\to1^+}\frac{x^2-1}{2x+4}=0$$ $$\lim_{x\to0^-}\frac{x^2-1}{2x+4}=-\frac{1}{4}$$
$$f(x)=\frac{x^2-1}{2x+4}=\frac{(x-1)(x+1)}{2(x+2)}$$ (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) $$\lim_{x\to0^-}f(x)=\lim_{x\to0^-}\frac{x^2-1}{2x+4}$$ $$\lim_{x\to0^-}f(x)=\frac{0^2-1}{2\times0+4}=-\frac{1}{4}$$