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

For $f(x)$ to be continuous at the origin, we only need to extend $f(x)$ to cover the point $(0,1)$.
$$f(x)=\frac{\tan(\tan x)}{\sin(\sin x)}$$ 1) First, find $\lim_{x\to0}\frac{\tan(\tan x)}{\sin(\sin x)}$ $$A=\lim_{x\to0}\frac{\tan(\tan x)\times\tan x}{\tan x}\times\lim_{x\to0}\frac{\sin x}{\sin(\sin x)\times\sin x}$$ $$A=\lim_{x\to0}\frac{\tan(\tan x)}{\tan x}\times\lim_{x\to0}\frac{\sin x}{\sin(\sin x)}\times\lim_{x\to0}\frac{\tan x}{\sin x}$$ $$A=J\times K\times L$$ - Consider $J$: Take $w=\tan x$. Then as $x\to0$, we would have $w=\tan x\to\tan0=0$ $$J=\lim_{w\to0}\frac{\tan w}{w}=\lim_{w\to0}\frac{\sin w}{w\cos w}=\lim_{w\to0}\frac{\sin w}{w}\times\lim_{w\to0}\frac{1}{\cos w}$$ $$J=1\times\frac{1}{\cos0}=\frac{1}{1}=1$$ - Consider $K$: Take $v=\sin x$. Then as $x\to0$, we would have $v=\sin x\to\sin0=0$ $$K=\lim_{v\to0}\frac{v}{\sin v}=\lim_{v\to0}\frac{1}{\frac{\sin v}{v}}=\frac{1}{1}=1$$ - Consider $L$: $$L=\lim_{x\to0}\frac{\tan x}{\sin x}=\lim_{x\to0}\frac{\sin x}{\cos x\times\sin x}=\lim_{x\to0}\frac{1}{\cos x}$$ $$L=\frac{1}{\cos0}=\frac{1}{1}=1$$ Therefore, $$A=1\times1\times1=1$$ 2) So here we see that $f(x)$ is not defined as $\sin x=0$ or $x=0$; in other words, $f(x)$ is not continuous at $x=0$. Yet we already have $\lim_{x\to0}\frac{\tan(\tan x)}{\sin(\sin x)}=1$, which means $f(x)$ approaches $1$ as $x\to0$. Therefore, for $f(x)$ to be continuous at the origin, we only need to extend $f(x)$ to cover the point $(0,1)$.