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.

Answer

I thought it might be instructive to present a way forward that goes back to "basics." Herein, we rely only on elementary inequalities and the squeeze theorem. To that end, we proceed with a primer.

PRIMER ON A SET OF ELEMENTARY INEQUALITIES:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}} \tag 1$$

for $x<1$.

And in THIS ANSWER, I showed using only elementary inequalities from geometry that the arctangent function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{|x|}{\sqrt{1+x^2}}\le |\arctan(x)|\le |x|} \tag 2$$

for all $x$.

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Using $(1)$ and $(2)$ we can write for $1>x>0$

$$\frac{x}{\sqrt{1+x^2}\left(\frac{2x}{1-2x}\right)}\le \frac{\arctan(x)}{e^{2x}-1}\le \frac{x}{2x} \tag 3$$

whereupon applying the squeeze theorem to $(3)$, we find that

$$\lim_{x\to 0^+}\frac{\arctan(x)}{e^{2x}-1}=\frac12$$

Similarly, using $(1)$ and $(2)$ for $x<0$ we can write

$$\frac{x}{\left(\frac{2x}{1-2x}\right)}\le \frac{\arctan(x)}{e^{2x}-1}\le \frac{x}{\sqrt{1+x^2}\,\left(2x\right)} \tag 4$$

whereupon applying the squeeze theorem to $(4)$, we find that

$$\lim_{x\to 0^-}\frac{\arctan(x)}{e^{2x}-1}=\frac12$$

Inasmuch as the limits from the right and left sides are equal we can conclude that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{\arctan(x)}{e^{2x}-1}=\frac12}$$