Answer
$$\lim_{x\to0}\tan\Big(\frac{\pi}{4}\cos(\sin x^{1/3})\Big)=1$$
The function is continuous at $x=0$
Work Step by Step
*Recall Theorem 10:
If $g$ is continuous at $b$ and $\lim_{x\to c}f(x)=b$, then $$\lim_{x\to c}g(f(x))=g(b)=g(\lim_{x\to c}f(x))$$
$$A=\lim_{x\to0}f(x)=\lim_{x\to0}\tan\Big(\frac{\pi}{4}\cos(\sin x^{1/3})\Big)$$
- First, consider the function $y=\tan x=\sin x/\cos x$
As $\cos x\ne0$, or $x\ne\pi/2+k\pi$ $(k\in Z)$, $\lim_{x\to c}\tan x= \tan c$
So function $y=\tan x$ is continuous for all $x\ne\pi/2+k\pi$ $(k\in Z)$
- For function $y=\cos x$, for all $x\in R$, $\lim_{x\to c}\cos x=\cos c$, so $y=\cos x$ is continuous on $(-\infty,\infty)$
That means we can apply Theorem 10 here as long as $\lim_{x\to0}\Big(\frac{\pi}{4}\cos(\sin x^{1/3})\Big)\ne\pi/2+k\pi$. In detail,
$$A=\tan\Big(\frac{\pi}{4}\cos(\lim_{x\to0}\sin x^{1/3})\Big)$$ $$A=\tan\Big(\frac{\pi}{4}\cos(\sin0^{1/3})\Big)$$ $$A=\tan\Big(\frac{\pi}{4}\cos0\Big)$$ $$A=\tan\Big(\frac{\pi}{4}\times1\Big)=\tan\frac{\pi}{4}=1$$
- Examine $f(0)$: $$\tan\Big(\frac{\pi}{4}\cos(\sin0^{1/3})\Big)=\tan\Big(\frac{\pi}{4}\cos0\Big)=\tan\Big(\frac{\pi}{4}\times1\Big)=\tan\frac{\pi}{4}=1$$
As $\lim_{x\to0}f(x)=f(0)$, the function is continuous at $x=0$
