calculus - Evaluation of a simple limit with Taylor Series

I would like to evaluate

$$\lim_{x\to0} \frac{e^{\sin x} - \sin^2x - 1}{x}$$
using Taylor Series expansion in a completely rigorous way.
What would a rigorous version of the following argument look like?

From

$$e^{\sin x} -1\sim_0 e^x-1 \sim_0 x \text{ and }\sin^2x \sim_0 x^2$$
we can find the limit
$$\lim_{x\to0} \frac{e^{\sin x} - \sin^2x - 1}{x} = \lim_{x\to0} \frac{x-\frac{x^2}{2}}{x} = 1.$$

In particular I'm not exactly sure how to deal with Landau symbols $o$ and $O$ in nested functions.

Answer

$\sin(x) = x + o(x)$, then $\sin^2(x) = x^2 + 2xo(x) + (o(x))^2 = x^2 + o(x)$, since

$$\lim_{x\to 0}\dfrac{2xo(x)+ (o(x))^2}{x} = \lim_{x\to 0}(2o(x)+(\dfrac{o(x)}{x})^2x = 0$$

$e^{\sin(x)} = 1+ \sin(x) + o(\sin(x)) = 1 + x + o(x) + o(x+ o(x)) = 1+ x + o(x)$, since

$$\lim_{x\to 0}\dfrac{o(x)+ o(x + o(x))}{x} = \lim_{x\to 0}\dfrac{o(x)}{x} + \lim_{x\to 0}\dfrac{o(x + o(x))}{x + o(x)}\dfrac{x+o(x)}{x} = 0$$

Thus $e^{\sin(x)} - \sin^2(x) -1 = x - x^2 + o(x)$