Solve the limit $\lim\limits_{x\to 0}\left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right)$ without using L'Hopital (with my attempts) Help me solve this limit, using simple limit work: $$\lim_{x\to 0}\left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right)$$
I tried extracting $\frac {e^x}{x}$ but that was dead end, then $\frac {e^x-1}{x}+\frac{1}{x}-\frac{1}{x} \frac{x}{\arctan{x}}$ but I got stuck there. 

Everything is locally a power series, so,
as $x \to 0$,
$\begin{array}\\
\left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right)
&=\left(\frac {1+x+O(x^2)}{x}-\frac {1}{x-x^3/3+O(x^5)}\right)\\
&=\left(\frac {(1+x+O(x^2))(1-x^2/3+O(x^4))-1}{x(1-x^2/3+O(x^4))}\right)\\
&=\left(\frac {(1+x+O(x^2))-1}{x(1+O(x^2))}\right)\\
&=\left(\frac {x+O(x^2)}{x+O(x^3))}\right)\\
&=1+O(x)\\
\end{array}
$
