<s>[INST] 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.
 [/INST] We have $$\lim_{x\to0}(1+\sin x)^{\dfrac1x}\cdot\lim_{x\to0}\left(1+\frac{\sin^2x}{1+\sin x}\right)^{\dfrac1x}$$
$$\lim_{x\to0}(1+\sin x)^{\dfrac1x}=\left(\lim_{x\to0}(1+\sin x)^{\frac1{\sin x}}\right)^{\lim_{x\to0}\dfrac{\sin x}x}=e^1$$
$$\lim_{x\to0}\left(1+\frac{\sin^2x}{1+\sin x}\right)^{\dfrac1x}=\left(\lim_{x\to0}\left(1+\frac{\sin^2x}{1+\sin x}\right)^{\dfrac{1+\sin x}{\sin^2x}}\right)^{\lim_{x\to0}\dfrac{\sin^2x}{x(1+\sin x)}}=e^0$$
As $\displaystyle\lim_{h\to0}\left(1+h\right)^{\frac1h}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$
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