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)

Question

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.

Answer 1

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}} $ $\sim$ $\Large{x^2\over x^2}$ $=$ $1$

Answer 2

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} $

Answer 3

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 $0$.