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

$y=\frac{1}{|x|+1}-\frac{x^2}{2}$ is continuous on $(-\infty,\infty)$
$$y=\frac{1}{|x|+1}-\frac{x^2}{2}$$ - Domain: because $|x|\ge0$, $|x|+1\gt0$, so the function is defined on $(-\infty,\infty)$ 1) Examine $f(x)=\frac{1}{|x|+1}$: As proved above, $|x|+1\ne0$ for all $x\in(-\infty,\infty)$. Therefore, $\lim_{x\to c}\frac{1}{|x|+1}=\frac{1}{|c|+1}$ for all $x\in(-\infty,\infty)$ So the function $f(x)=\frac{1}{|x|+1}$ is continuous on $(-\infty,\infty)$. 2) Examine $g(x)=\frac{x^2}{2}$ $\lim_{x\to c}\frac{x^2}{2}=\frac{c^2}{2}$ for all $x\in(-\infty,\infty)$ So the function $g(x)=\frac{x^2}{2}$ is continuous on $(-\infty,\infty)$. According to Theorem 8, the subtraction of two continous function at $x=c$ is also continuous at $x=c$. Therefore, $y=\frac{1}{|x|+1}-\frac{x^2}{2}$ is continuous on $(-\infty,\infty)$