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

$$\lim_{x\to0}\tan\Big(\frac{\pi}{4}\cos(\sin x^{1/3})\Big)=1$$ The function is continuous at $x=0$
*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 $x=0$