$$\lim_{x \to 0} \frac{e^{2\sin x} - 2\sin x - 1}{x^2}$$

We need $$\displaystyle\lim_{x\to 0}\frac{e^{2\sin x}-2\sin x-1}{x^{2}}$$.

Step 1 : Maclaurin series of the basic functions
For any small $$t$$,
$$e^{t}=1+t+\frac{t^{2}}{2}+\frac{t^{3}}{6}+O(t^{4})$$
$$\sin t=t-\frac{t^{3}}{6}+O(t^{5})$$

Step 2 : Expand $$\sin x$$ and $$2\sin x$$
$$\sin x=x-\frac{x^{3}}{6}+O(x^{5})$$ $$-(1)$$
Multiply by $$2$$:
$$2\sin x=2x-\frac{x^{3}}{3}+O(x^{5})$$ $$-(2)$$

Step 3 : Expand $$e^{2\sin x}$$
Put $$t=2\sin x$$ of $$(2)$$ into the exponential series:
$$e^{2\sin x}=1+\bigl(2\sin x\bigr)+\frac{(2\sin x)^{2}}{2}+O\!\bigl((2\sin x)^{3}\bigr)$$ $$-(3)$$

Step 4 : Form the numerator
Subtract $$1+2\sin x$$ from $$(3)$$:
$$e^{2\sin x}-2\sin x-1=\frac{(2\sin x)^{2}}{2}+O\!\bigl((2\sin x)^{3}\bigr)$$
$$=\;2\sin^{2}x+O(x^{3})$$ $$-(4)$$

Step 5 : Expand $$\sin^{2}x$$ up to $$x^{2}$$
From $$(1)$$,
$$\sin^{2}x=\Bigl(x-\frac{x^{3}}{6}\Bigr)^{2}+O(x^{5})$$
$$=x^{2}-\frac{x^{4}}{3}+O(x^{5})$$ $$-(5)$$

Step 6 : Substitute $$(5)$$ into $$(4)$$
$$e^{2\sin x}-2\sin x-1=2\bigl(x^{2}-\frac{x^{4}}{3}\bigr)+O(x^{5})$$
$$=2x^{2}+O(x^{4})$$ $$-(6)$$

Step 7 : Divide by $$x^{2}$$ and take the limit
$$\frac{e^{2\sin x}-2\sin x-1}{x^{2}}=\frac{2x^{2}+O(x^{4})}{x^{2}}$$
$$=2+O(x^{2})$$.

As $$x\to 0$$, the error term $$O(x^{2})\to 0$$, so

$$\displaystyle\lim_{x\to 0}\frac{e^{2\sin x}-2\sin x-1}{x^{2}}=2$$.

The limit equals $$2$$ ⇒ Option D.