Evaluating $\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}$ I did this:
$$\begin{align}
\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x} &\sim \lim_{x \to 0} \frac{(1 + x + x^2)^{1/x} - (1 + x)^{1/x}}{x} = \\
&= \lim_{x \to 0} \left [ (1+x)^{1/x} \frac{\left ( \frac{1 + x + x^2}{1 + x} \right )^{1/x} - 1}{x} \right ] =\\
&= \lim_{x \to 0} (1+x)^{1/x} \cdot \lim_{x \to 0} \frac{\left ( 1 + \frac{x^2}{1 + x} \right )^{1/x} - 1}{x} =\\
&= e \cdot \lim_{x \to 0} \frac{e^{\frac{1}{x} \cdot \ln \left ( 1 + \frac{x^2}{1+x} \right )} - 1}{x} \sim \\
&\sim e \cdot \lim_{x \to 0} \frac{e^{\frac{x}{1 + x}} - 1}{x} \sim \\
&\sim e \cdot \lim_{x \to 0} \frac{1}{1 + x} = e
\end{align}$$
Is it right? If it is, how to evaluate the limit faster? It was pretty long the way I did it.

No the most rigorous, but fast. Define $n=\frac 1x$, and assume it integer (keep in mind the dependency on $x$).
$$\frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}=\frac{(1 + \sin x + \sin^2 x)^n - (1 + \sin x)^n}{x}.$$
Factoring the numerator as a binomial of the form $a^n-b^n$, you will get two factors:


*

*the difference of $a-b$, i.e. $\sin^2x$,

*the sum of products of powers $a^{n-1-k}b^k$. All these terms behave like $(1+x+o(x^2))^{n-1}$ and there are $n$ of them. You can recognize the definition of $e$.
So you end up with
$$\frac{\sin^2x}xne$$
which tends to $e$.
