Limit involving exponentials and arctangent without L'Hôpital $$\lim_{x\to0}\frac{\arctan x}{e^{2x}-1}$$
How to do this without L'Hôpital and such? $\arctan x=y$, then we rewrite it as $\lim_{y\to0}\frac y{e^{2\tan y}-1}$, but from here I'm stuck.

Another way using Taylor series $$\tan^{-1}(x)=x-\frac{x^3}{3}+O\left(x^4\right)$$ $$e^{2x}-1=2 x+2 x^2+\frac{4 x^3}{3}+O\left(x^4\right)$$ $$\frac{\tan^{-1}(x) } {e^{2x}-1 }=\frac{x-\frac{x^3}{3}+O\left(x^4\right) } {2 x+2 x^2+\frac{4 x^3}{3}+O\left(x^4\right) }=\frac{1-\frac{x^2}{3}+O\left(x^3\right) } {2 +2 x+\frac{4 x^2}{3}+O\left(x^3\right) }$$ Performing the long division $$\frac{\tan^{-1}(x) } {e^{2x}-1 }=\frac{1}{2}-\frac{x}{2}+O\left(x^3\right)$$ which shows the limit and how it is approached.
