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. 
 A: 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}
$
A: Rewrite everything as:
$ \Large{e^x\arctan x -x\over x \arctan x} $
$ \arctan x $ is asymptotically equal to $x$ as $x \to 0$ so we have:
$ \Large{e^x\arctan x -x\over x \arctan x}$ $\sim$ $\Large{{xe^x -x}\over{x^2}} $
Group by $x$ the numerator:
$\Large{{x(e^x -1)}\over{x^2}} $
$ e^x -1 $ is asymptotically equivalent to $x$ as $x \to 0$ so we have:
$\Large{{x(e^x -1)}\over{x^2}} $ $\sim$ $\Large{x^2\over x^2}$ $=$ $1$
A: Your own approach works fine. Just split the expression under limit as $$\frac{e^x-1}{x}+\frac{\arctan x-x} {x\arctan x} $$ The first fraction tends to $1$ and one show that the second fraction tends to $0$ so that the desired limit is $1$. Since $(\arctan x) /x\to 1$ the limit of second fraction above is same as that of $$\frac{\arctan x-x} {(\arctan x) ^2}$$ Putting $x=\tan t$ we see that $t\to 0$ and the above fraction reduces to $$\frac{t-\tan t} {t^2}$$ Clearly the limit of above expression is same as that of $$\frac{t\cos t-\sin t} {t^2}=\frac{\cos t - 1}{t}+\frac{t-\sin t} {t^2}$$ and you can easily prove that these two fractions tend to $0$.
