Limit of $\arctan(x)/x$ as $x$ approaches $0$? Quick question:
I came across the following limit: $$\lim_{x\rightarrow 0^{+}}\frac{\arctan(x)}{x}=1.$$ 
It seems like the well-known limit: 
$$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1.$$ 
Can anyone show me how to prove it?

Recall (see the diagram below)  that for $0\le t<{\pi\over2}$:
$$\tag{1}
\sin t \le t \le \tan t.
$$
Taking $t =\arctan x$ in $(1)$,  we have, for $x>0$:
$$
\sin\bigl(\arctan(x)\bigr)\le \arctan(x)\le x.
$$
But
$$
\sin\bigl(\arctan (x)\bigr) ={x\over \sqrt{1+x^2}};
$$
whence, for $x>0$:
$$
{x\over \sqrt{1+x^2}}\le \arctan(x)\le x.
$$
So, for $x>0$, we have
$$
{1\over \sqrt{1+x^2}}\le {\arctan(x)\over x}\le 1;
$$
and it follows from the Squeeze Theorem that
$$
\lim_{x\rightarrow0^+} {\arctan(x)\over x}=1.
$$




