Evaluating $\lim_{x \to 0} \dfrac{\arctan x}{x}$ 
Evaluate $$\lim_{x \to 0} \dfrac{\arctan x}{x}$$

I know that it's $1$ using L'Hopital's rule or numerically, but what about algebraically? Thanks in advance and I really appreciate it.
 A: With $y=\tan x$,
$$\lim_{x\to 0}\frac{\arctan x}{x}=\lim_{y\to 0}\frac{y}{\tan y}=\lim_{y\to 0}\frac{y\cos y}{\sin y}=\lim_{y\to 0}\cos y\cdot \lim_{y\to 0}\frac y{\sin y}=\lim_{y\to 0}\frac y{\sin y}=1$$
A: Use the power series expansion of $\arctan$, $\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\ldots$, to get $\arctan(x)/x=1-\frac{x^2}{3}+\frac{x^4}{5}-\ldots$.
A: You can use the series expansion at $x = 0$
$$\dfrac{\arctan x}{x} = 1 - \dfrac{x^2}{3} + \dfrac{x^4}{5} + O(x^6)$$
Thus
$$\lim_{x \to 0} \dfrac{\arctan x}{x} = \lim_{x \to 0} {1 - \dfrac{x^2}{3} + \dfrac{x^4}{5} + O(x^6)} = 1 $$
A: $\lim_{x\to 0}\dfrac{\arctan x}{x}=\lim_{x\to 0}\dfrac{\arctan x-\arctan 0}{x-0}=\dfrac{d\arctan(x)}{dx}(0)=\cos^2(\arctan 0)=1.$
A: If you do not want to use L'Hopital, or series expansion, then I suggest the following :
Notice that for any $ x\in\mathbb{R}^{*} $, we have : $$ \frac{\arctan{x}}{x}=\int_{0}^{1}{\frac{\mathrm{d}y}{1+x^{2}y^{2}}}=1-x^{2}\int_{0}^{1}{\frac{y^{2}}{1+x^{2}y^{2}}\,\mathrm{d}y} $$
Since, for all $ x\in\mathbb{R}^{*} $, $$ \left|x^{2}\int_{0}^{1}{\frac{y^{2}}{1+x^{2}y^{2}}\,\mathrm{d}y}\right|\leq x^{2}\int_{0}^{1}{y^{2}\,\mathrm{d}y}=\frac{x^{2}}{3}\underset{x\to 0}{\longrightarrow}0 $$
Then $ \lim\limits_{x\to 0}{x^{2}\int_{0}^{1}{\frac{y^{2}}{1+x^{2}y^{2}}\,\mathrm{d}y}}=0 $, which means : \begin{aligned} \lim_{x\to 0}{\frac{\arctan{x}}{x}}&=\lim_{x\to 0}{\left(1-x^{2}\int_{0}^{1}{\frac{y^{2}}{1+x^{2}y^{2}}\,\mathrm{d}y}\right)}\\ &=1-0\\ \lim_{x\to 0}{\frac{\arctan{x}}{x}}&=1\end{aligned}
