Answer
$$\lim_{t\to0}\sin(\frac{\pi}{2}\cos(\tan t))=1$$
The function is continuous at $t=0$
Work Step by Step
*Recall Theorem 10:
If $g$ is continuous at $b$ and $\lim_{x\to c}f(x)=b$, then $$\lim_{x\to c}g(f(x))=g(b)=g(\lim_{x\to c}f(x))$$
$$A=\lim_{t\to0}f(t)=\lim_{t\to0}\sin(\frac{\pi}{2}\cos(\tan t))$$
For the domain $(-\infty,\infty)$, $\lim_{x\to c}\sin x= \sin c$ and $\lim_{x\to c}\cos x= \cos c$
So both functions $y=\sin x$ and $y=\cos x$ are continuous on $(-\infty,\infty)$
That means we can apply Theorem 10 here. In detail,
$$A=\sin(\frac{\pi}{2}\lim_{t\to0}\cos(\tan t))$$
$$A=\sin(\frac{\pi}{2}\cos(\lim_{t\to0}\tan t))$$
$$A=\sin(\frac{\pi}{2}\cos(\tan0))$$
$$A=\sin(\frac{\pi}{2}\cos0)$$
$$A=\sin(\frac{\pi}{2}\times1)$$
$$A=\sin\frac{\pi}{2}=1$$
- Examine $f(0)$: $$\sin(\frac{\pi}{2}\cos(\tan0))=\sin(\frac{\pi}{2}\cos0)=\sin(\frac{\pi}{2}\times1)=\sin\frac{\pi}{2}=1$$
As $\lim_{t\to0}f(t)=f(0)$, the function is continuous at $t=0$
