## Calculus: Early Transcendentals 8th Edition

1) Check the continuity of $f(x)=\sin x$ 2) Check the continuity of $f(x)=\cos x$ 3) Check the continuity of $f(x)$ at $\frac{\pi}{4}$
$\sin x$ is continuous on $R$, so it is continuous on $(-\infty,\frac{\pi}{4})$ $\cos x$ is continuous on $R$, so it is continuous on $(\frac{\pi}{4},\infty)$ Therefore, $f(x)$ is continuous on $(-\infty,\frac{\pi}{4})\cup(\frac{\pi}{4},\infty)$ Now we only need to check if $f(x)$ is continuous at $\frac{\pi}{4}$. By definition, $f(x)$ is continuous at $\frac{\pi}{4}$ if and only if $$\lim\limits_{x\to\frac{\pi}{4}}f(x)=f(\frac{\pi}{4})$$ 1) For $x\frac{\pi}{4}$, we have $\lim\limits_{x\to(\frac{\pi}{4})^+}f(x)=\lim\limits_{x\to(\frac{\pi}{4})^+}\cos x=\cos(\frac{\pi}{4})=\frac{\sqrt2}{2}$ Since $\lim\limits_{x\to(\frac{\pi}{4})^-}f(x)=\lim\limits_{x\to(\frac{\pi}{4})^+}f(x)=\frac{\sqrt2}{2}$ Therefore, $\lim\limits_{x\to\frac{\pi}{4}}f(x)=\frac{\sqrt2}{2}$ Notice that $f(\frac{\pi}{4})=\cos(\frac{\pi}{4})=\frac{\sqrt2}{2}$ Since $\lim\limits_{x\to\frac{\pi}{4}}f(x)=f(\frac{\pi}{4})$, $f(x)$ is continuous at $\frac{\pi}{4}$. Overall, $f(x)$ is continuous on $(-\infty,\infty)$.