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

$$\lim_{x\to2^+}\frac{1}{x^2-4}=\infty$$ $$\lim_{x\to2^-}\frac{1}{x^2-4}=-\infty$$ $$\lim_{x\to-2^+}\frac{1}{x^2-4}=-\infty$$ $$\lim_{x\to-2^-}\frac{1}{x^2-4}=\infty$$
$$f(x)=\frac{1}{x^2-4}=\frac{1}{(x-2)(x+2)}$$ (a) As $x\to2^+$, $(x-2)\to0^+\gt0$, while $(x+2)\to4^+\gt0$. $(x-2)(x+2)\gt0$ as a result. That means $\frac{1}{(x-2)(x+2)}$ will approach $\infty$. In other words, $$\lim_{x\to2^+}\frac{1}{x^2-4}=\infty$$ (b) As $x\to2^-$, $(x-2)\to0^-\lt0$, while $(x+2)\to4^-\gt0$. $(x-2)(x+2)\lt0$ as a result. That means $\frac{1}{(x-2)(x+2)}$ will approach $-\infty$. In other words, $$\lim_{x\to2^-}\frac{1}{x^2-4}=-\infty$$ (c) As $x\to-2^+$, $(x+2)\to0^+\gt0$, while $(x-2)\to-4^+\lt0$. $(x-2)(x+2)\lt0$ as a result. That means $\frac{1}{(x-2)(x+2)}$ will approach $-\infty$. In other words, $$\lim_{x\to-2^+}\frac{1}{x^2-4}=-\infty$$ (d) As $x\to-2^-$, $(x+2)\to0^-\lt0$, while $(x-2)\to-4^-\lt0$. $(x-2)(x+2)\gt0$ as a result. That means $\frac{1}{(x-2)(x+2)}$ will approach $\infty$. In other words, $$\lim_{x\to-2^-}\frac{1}{x^2-4}=\infty$$