_ x → 0 e ^ 2| x| - 2| x| - 1 x ^ 2

Given, $\lim_{x \to 0}\dfrac{{e}^{2\left|\sin x\right|} - 2\left|\sin x\right| - 1}{{x}^{2}}$

Now, using the expansion of ${e}^{x} = 1 + \dfrac{x}{1 !} + \dfrac{{x}^{2}}{2 !} + . . . . . . . . \infty$ we get,

$= \lim_{x \to 0}\dfrac{\left(1 + \dfrac{\left|2\sin x\right|}{1 !} + \dfrac{{\left|2\sin x\right|}^{2}}{2 !} + . . . . . \infty\right) - \left|2\sin x\right| - 1}{{x}^{2}}$

$= \lim_{x \to 0}\dfrac{\dfrac{{\left|2\sin x\right|}^{2}}{2 !} + \dfrac{{\left|2\sin x\right|}^{3}}{3 !} . . . . . \infty}{{x}^{2}}$

$= \lim_{x \to 0}\dfrac{2\sin^{2}x + \dfrac{{\left|2\sin x\right|}^{3}}{3 !} . . . . . \infty}{{x}^{2}}$

$= 2\left\{\text{as}\lim_{x \to 0}\dfrac{\sin^{2}x}{{x}^{2}} = 1\text{and other terms will become zero}\right\}$