Answer
a) $f(x)$ is continuous for all $x\ne\frac{\pi}{2}+k\pi$ $(k\in Z)$
b) $g(x)$ is continuous for all $x\ne k\pi$ $(k\in Z)$.
c) $h(x)$ is continuous on $(-\infty,\pi)\cup(\pi,\infty)$.
d) $k(x)$ is continuous on $(-\infty,0)\cup(0,\infty)$.
Work Step by Step
a) $f(x)=\tan x=\frac{\sin x}{\cos x}$
- Domain: $f(x)$ is defined where $\cos x\ne0$, which means $x\ne\frac{\pi}{2}+k\pi$ $(k\in Z)$
- For all $x=c$ in its domain, we have $$\lim_{x\to c}f(x)=\frac{\sin c}{\cos c}=f(c)$$
- Since $f(x)$ is not defined as $x=\frac{\pi}{2}+k\pi$, $f(x)$ is not continuous for all the values $x=\frac{\pi}{2}+k\pi$
Therefore, $f(x)$ is continuous on its domain, which is $R$ except $x=\frac{\pi}{2}+k\pi$ $(k\in Z)$
b) $g(x)=\csc x=\frac{1}{\sin x}$
- Domain: $g(x)$ is defined where $\sin x\ne0$, which means $x\ne k\pi$ $(k\in Z)$
- For all $x\ne k\pi$, we have $$\lim_{x\to c}g(x)=\frac{1}{\sin c}=f(c)$$
- Since $g(x)$ is not defined for all $x=k\pi$, $g(x)$ is not continuous for all $x=k\pi$ as well.
Therefore, $g(x)$ is continuous on its domain, which is $R$ except $x=k\pi$ $(k\in Z)$.
c) $h(x)=\frac{\cos x}{x-\pi}$
- Domain: $(-\infty,\pi)\cup(\pi,\infty)$
- For all $x=c\in (-\infty,\pi)\cup(\pi,\infty)$, we have $$\lim_{x\to c}h(x)=\frac{\cos c}{c-\pi}=f(c)$$
- For $x=\pi$: Since $h(\pi)$ is not defined, $h(x)$ is not continuous at $x=\pi$.
Therefore, $h(x)$ is continuous on its domain, which is $(-\infty,\pi)\cup(\pi,\infty)$.
d) $k(x)=\frac{\sin x}{x}$
- Domain: $(-\infty,0)\cup(0,\infty)$
- For all $x=c\in (-\infty,0)\cup(0,\infty)$, we have $$\lim_{x\to c}k(x)=\frac{\sin c}{c}=f(c)$$
- For $x=0$: Since $k(x)$ is not defined as $x=0$, $k(x)$ is not continuous at $x=0$
Therefore, $k(x)$ is continuous on its entire domain, which is $(-\infty,0)\cup(0,\infty)$.
