552. Find the limits:
1) $\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}\right)^{x}$
2) $\lim _{n \rightarrow \infty}\left(\frac{n-1}{n+2}\right)^{2 n+1}$
3) $\lim _{x \rightarrow 0} \sqrt[x]{1+3 x}$
4) $\lim _{x \rightarrow 0}(\cos 2 x)^{\frac{1}{\sin ^{2} x}}$
Solution. We have an indeterminate form of $1^{\infty}$. Each of the given expressions under the limit sign will be transformed so that the problem reduces to the second remarkable limit.

1) $\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}\right)^{x}=\lim _{x \rightarrow \infty}\left[\left(1+\frac{a}{x}\right)^{\frac{x}{a}}\right]^{\frac{a}{x} \cdot x}=e^{a}$, since $\lim _{x \rightarrow \infty}(1+$ $\left.+\frac{a}{x}\right)^{\frac{x}{a}}=\lim _{\alpha \rightarrow 0}(1+\alpha)^{\frac{1}{\alpha}}=e$, where $\alpha=\frac{a}{x}$, and the exponent $\frac{a}{x} \cdot x=a$.
2) $\lim _{n \rightarrow \infty}\left(\frac{n-1}{n+2}\right)^{2 n+1}=\lim _{n \rightarrow \infty}\left(1+\frac{n-1}{n+2}-1\right)^{2 n+1}=$ $=\lim _{n \rightarrow \infty}\left(1+\frac{-3}{n+2}\right)^{2 n+1}=\lim _{n \rightarrow \infty}\left[\left(1+\frac{-3}{n+2}\right)^{\frac{n+2}{-3}}\right]^{\frac{-3}{n+2} \cdot(2 n+1)}=$ $=e^{-6}=\frac{1}{e^{6}}$, since $\lim _{n \rightarrow \infty}\left(1+\frac{-3}{n+2}\right)^{\frac{n+2}{-3}}=e$ by the formula and $\lim _{n \rightarrow \infty} \frac{-3(2 n+1)}{n+2}=-6$.
3) $\lim _{x \rightarrow 0} \sqrt[x]{1+3 x}=\lim _{x \rightarrow 0}(1+3 x)^{\frac{1}{x}}=\lim _{x \rightarrow 0}\left[(1+3 x)^{\frac{1}{3 x}}\right]^{3 x \cdot \frac{1}{x}}=e^{3}$, since $\lim _{x \rightarrow 0}(1+3 x)^{\frac{1}{3 x}}=\lim _{x \rightarrow 0}(1+\alpha)^{\frac{1}{\alpha}}=e$, where $\alpha=3 x$, and the exponent $3 x \cdot \frac{1}{x}=3$.
4) Here we also have an indeterminate form of $1^{\infty}$:

$$
\begin{aligned}
& \lim _{x \rightarrow 0}(\cos 2 x)^{\frac{1}{\sin ^{2} x}}=\lim _{x \rightarrow 0}(1+\cos 2 x-1)^{\frac{1}{\sin ^{2} x}}= \\
= & \lim _{x \rightarrow 0}\left\{\left[1+\left(-2 \sin ^{2} x\right)\right]^{\frac{-1}{2 \sin ^{2} x}}\right\}^{\frac{-2 \sin ^{2} x}{\sin ^{2} x}}=e^{-2}=\frac{1}{e^{2}}
\end{aligned}
$$

We used the formula $1-\cos 2 x=2 \sin ^{2} x$ and the second remarkable limit.

Find the limits: