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?
 A: Since $\lim\limits_{x\to0}\arctan(x)=0$, letting $x=\tan(\theta)$ yields
$$
\lim_{x\to0}\frac{\arctan(x)}{x}=\lim_{\theta \to0}\frac{\theta}{\tan(\theta)}\tag{1}
$$
and $(1)$ is shown to be $\frac11$ in equation $(5)$ of this answer.
A: We can make use of L'Hopital's rule. Since $\frac{d}{dx}\arctan x=\frac{1}{x^2+1}$ and $\frac{d}{dx}x=1$, we have
$$\lim\limits_{x\to0^+}\frac{\arctan x}{x}=\lim\limits_{x\to0^+}\frac{1}{x^2+1}=1.$$
A: 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
$$
\lim_{x\rightarrow0^+} {\arctan(x)\over x}=1.
$$




A: $$\lim_{x\rightarrow 0^{+}}\frac{\arctan(x)}{x}= \lim_{h\rightarrow 0^{+}}\frac{\arctan(0+h) -\arctan(0)}{h} = \arctan'(0) = \frac{1}{1+0^2} = 1$$
A: If you don't yet have access (which is often the case) to such relatively advanced tools as derivatives, L'Hopital's rule, and series expansion, here is a very simple proof:
Once you know:
$$\lim_{x \rightarrow 0} \frac{\sin x}{x}= 1$$
You can prove that
$$\lim_{x \rightarrow 0} \frac{\tan x}{x}= 1$$
Indeed,
$$\lim_{x \rightarrow 0} \frac{\tan x}{x}= \lim_{x \rightarrow 0} \frac{\sin x}{x \cdot \cos x}= \lim_{x \rightarrow 0} \frac{\sin x}{x} \lim_{x \rightarrow 0} \frac{1}{\cos x}= 1\cdot1 = 1$$
Now you make a simple substitution:
$$t = \arctan x \implies x = \tan t$$
$$x \rightarrow 0 \implies t \rightarrow 0$$
Finally,
$$\lim_{x \rightarrow 0} \frac{\arctan x}{x} = \lim_{t \rightarrow 0} \frac{t}{\tan t} = 1$$
(the last limit equals $1$, as proved above).
If you were actually looking for the proof $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ then there are plenty of nice unit circle proofs on the internet. Maybe you could try this one.
A: Can you do the Taylor series' expansion of $\arctan (x)$? If you can, then it is easy to solve your limit. If you cannot, refer to this.
A: If you know that $\tan x \underset{x \rightarrow 0}{\sim} x $ you could compute : 
$x = \arctan(\tan x) \underset{x \rightarrow 0}{\sim} \arctan x$ and then $ \frac{\arctan x}{x} \underset{x \rightarrow 0}{\rightarrow} 1$
Edit : sorry I don't see the proof above which use the same idea.
