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

$y=\tan\frac{\pi x}{2}$ is continuous for all $x\ne1+2k$ $(k\in Z)$
$$y=\tan\frac{\pi x}{2}=\frac{\sin\frac{\pi x}{2}}{\cos\frac{\pi x}{2}}$$ - Domain: $y$ is defined where $\cos(\pi x/2)\ne0$, which means $$\frac{\pi }{2}x\ne \frac{\pi}{2}+k\pi$$ $$x\ne1+\frac{k\pi}{\frac{\pi}{2}}$$ $$x\ne1+2k(k\in Z)$$ - As $x$ approaches any values of $1+2k$, $\cos(\pi x/2)$ would approach $0$, thus $\frac{\sin(\pi x/2)}{\cos(\pi x/2)}$ will approach infinity and does not reach for any definite value. In other words, $\lim_{x\to(1+2k)}\frac{\sin(\pi x/2)}{\cos(\pi x/2)}$ does not exist, so the function $y=\tan(\pi x/2)$ is discontinuous at all points $x=1+2k$ $(k\in Z)$. So for all $x\ne1+2k$: $$\lim_{x\to c}\frac{\sin(\pi x/2)}{\cos(\pi x/2)}=\frac{\sin(\pi c/2)}{\cos(\pi c/2)}$$ Therefore, $y=\tan(\pi x/2)=\frac{\sin(\pi x/2)}{\cos(\pi x/2)}$ is continuous in the domain defined. In conclusion, $y=\tan\frac{\pi x}{2}$ is continuous for all $x\ne1+2k$ $(k\in Z)$