118. Find $\lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{\alpha}$.
Solution. We will transform this limit into the form (1). For this, we multiply the numerator and the denominator of the fraction by 2, and take the constant factor 2 outside the limit sign. We have

$$
\lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{\alpha}=\lim _{\alpha \rightarrow 0} \frac{2 \sin 2 \alpha}{2 \alpha}=2 \lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{2 \alpha}
$$

Considering that if $\alpha \rightarrow 0$, then $2 \alpha \rightarrow 0$, we get

$$
\lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{\alpha}=2 \lim _{2 \alpha \rightarrow 0} \frac{\sin 2 \alpha}{2 \alpha}=2 \lim _{x \rightarrow 0} \frac{\sin x}{x}=2 \cdot 1=2
$$

If $x \rightarrow \infty$, then the following relation holds

$$
\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e \approx 2.7182 \ldots
$$

which is known as the second remarkable limit. The proof of its validity is provided in detailed courses of mathematical analysis.

The number $e$ is of great importance in mathematics. Logarithms with base $e$ are called natural, and for them, the notation $\ln$ is used. Thus, $\ln x=\log _{e} x$. For example, $\ln 2 \approx 0.6931$.