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

$$\lim_{x\to1^+}\frac{x}{x^2-1}=\infty$$ $$\lim_{x\to1^-}\frac{x}{x^2-1}=-\infty$$ $$\lim_{x\to-1^+}\frac{x}{x^2-1}=\infty$$ $$\lim_{x\to-1^-}\frac{x}{x^2-1}=-\infty$$
$$f(x)=\frac{x}{x^2-1}=\frac{x}{(x-1)(x+1)}$$ (a) As $x\to1^+\gt0$, $(x-1)\to0^+\gt0$, while $(x+1)\to2^+\gt0$. $\frac{x}{(x-1)(x+1)}\gt0$ as a result. That means $\frac{x}{(x-1)(x+1)}$ will approach $\infty$. In other words, $$\lim_{x\to1^+}\frac{x}{x^2-1}=\infty$$ (b) As $x\to1^-\gt0$, $(x-1)\to0^-\lt0$, while $(x+1)\to2^-\gt0$. $\frac{x}{(x-1)(x+1)}\lt0$ as a result. That means $\frac{x}{(x-1)(x+1)}$ will approach $-\infty$. In other words, $$\lim_{x\to1^-}\frac{x}{x^2-1}=-\infty$$ (c) As $x\to-1^+\lt0$, $(x+1)\to0^+\gt0$, while $(x-1)\to-2^+\lt0$. $\frac{x}{(x-1)(x+1)}\gt0$ as a result. That means $\frac{x}{(x-1)(x+1)}$ will approach $\infty$. In other words, $$\lim_{x\to-1^+}\frac{x}{x^2-1}=\infty$$ (d) As $x\to-1^-\lt0$, $(x+1)\to0^-\lt0$, while $(x-1)\to-2^-\lt0$. $\frac{x}{(x-1)(x+1)}\lt0$ as a result. That means $\frac{x}{(x-1)(x+1)}$ will approach $-\infty$. In other words, $$\lim_{x\to-1^-}\frac{x}{x^2-1}=-\infty$$