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.

We have $$\begin{aligned}L &= \lim_{x \to 0}\frac{(1 + \sin x + \sin^{2}x)^{1/x} - (1 + \sin x)^{1/x}}{x}\\
&= \lim_{x \to 0}(1 + \sin x)^{1/x}\dfrac{\left(1 + \dfrac{\sin^{2}x}{1 + \sin x}\right)^{1/x} - 1}{x}\\
&= \lim_{x \to 0}e\cdot\dfrac{\left(1 + \dfrac{\sin^{2}x}{1 + \sin x}\right)^{1/x} - 1}{x}\\
&= e\lim_{x \to 0}\dfrac{\left(1 + \dfrac{\sin^{2}x}{1 + \sin x}\right)^{1/x} - 1}{x}\\
&= e\lim_{x \to 0}g(x) = eA\end{aligned}$$ The limit $A = \lim_{x \to 0}g(x)$ can be handled with elementary techniques. Let $$f(x) = \frac{\sin^{2}x}{1 + \sin x}$$ so that $f(x) \to 0$ as $x \to 0$, then we can see that $$\begin{aligned}g(x) &= \frac{(1 + f(x))^{1/x} - 1}{x}\\
&= \dfrac{\exp\left(\dfrac{\log(1 + f(x))}{x}\right) - 1}{x}\\
&= \dfrac{\exp(t) - 1}{x}\end{aligned}$$ Note that $$\begin{aligned}t &= \frac{\log(1 + f(x))}{x}\\
&= \frac{\log(1 + f(x))}{f(x)}\cdot\frac{f(x)}{x}\\
&= \frac{\log(1 + f(x))}{f(x)}\cdot\frac{\sin x}{x}\cdot\frac{\sin x}{1 + \sin x}\\
&\to 1\cdot 1\cdot 0 = 0\end{aligned}$$ Thus we can write $$\begin{aligned}A &= \lim_{x \to 0}g(x)\\
&= \lim_{x \to 0}\frac{e^{t} - 1}{t}\cdot\frac{t}{x}\\
&= \lim_{x \to 0}1\cdot\frac{\log(1 + f(x))}{x^{2}}\\
&= \lim_{x \to 0}\frac{\log (1 + f(x))}{f(x)}\cdot\frac{f(x)}{x^{2}}\\
&= \lim_{x \to 0}1\cdot\frac{\sin^{2}x}{x^{2}(1 + \sin x)}\\
&= \lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\frac{1}{1 + \sin x} = 1\cdot 1 = 1\end{aligned}$$ Thus $L = eA = e\cdot 1 = e$ and the final answer is $e$.
Note: We have used the the limit $(1 + \sin x)^{1/x} \to e$ as $x \to 0$. This is easily handled because $$\log(1 + \sin x)^{1/x} = \frac{\log(1 + \sin x)}{x} = \frac{\log(1 + \sin x)}{\sin x}\cdot\frac{\sin x}{x}\to 1\cdot 1 = 1$$
