88. 1) $\lim _{n \rightarrow \infty}\left(1+\frac{a}{n}\right)^{n}$
2) $\lim _{x \rightarrow 0} \sqrt[x]{1-2 x}$
3) $\lim _{t \rightarrow \infty}\left(\frac{t-3}{t+2}\right)^{2 t+1}$
4) $\lim (\tan x)^{\tan 2 x}$.
$x \rightarrow \frac{\pi}{4}$
Solution. First, we ensure that with the specified change in the argument, the function represents a power where the base tends to one and the exponent tends to infinity (case $1^{*}$). Then, we transform the function to use the 2nd remarkable limit.

1) Assuming $n=a x$, we get $x \rightarrow \infty$ when $n \rightarrow \infty$, and

$$
\begin{aligned}
\lim _{n \rightarrow \infty}\left(1+\frac{a}{n}\right)^{n} & =\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{a x}=\lim \left[\left(1+\frac{1}{x}\right)^{x}\right]^{a}= \\
& =\left[\lim \left(1+\frac{1}{x}\right)^{x}\right]^{a}=e^{a}
\end{aligned}
$$

Another solution without changing the variable is also possible:

$$
\lim _{n \rightarrow \infty}\left(1+\frac{a}{n}\right)^{n}=\operatorname{li:11}\left[\left(1+\frac{a}{n}\right)^{\frac{n}{a}}\right]^{a}=\left[\lim _{\frac{n}{a} \rightarrow \infty}\left(1+\frac{a}{n}\right)^{\frac{n}{a}}\right]^{a}=e^{a}
$$

2) Assuming $-2 x=\alpha$, we find $\alpha \rightarrow 0$ when $x \rightarrow 0$, and

$$
\lim _{x \rightarrow 0}(1-2 x)^{\frac{1}{x}}=\lim _{a \rightarrow 0}(1+\alpha)^{-\frac{2}{a}}=\left[\operatorname{lim11}(1+\alpha)^{\frac{1}{\alpha}}\right]^{-2}=e^{-2}
$$

3) Excluding the integer part from the fraction, we assume $-\frac{5}{t+2}=x$:

$$
\begin{gathered}
\lim _{t \rightarrow \infty}\left(\frac{t-3}{t+2}\right)^{2 t+1}=\lim \left(1-\frac{5}{t+2}\right)^{2 t+1}=\lim _{x \rightarrow 0}(1+x)^{-\frac{10}{x}-3}= \\
=\left[\lim (1+x)^{\frac{1}{x}}\right]^{-10} \cdot \lim (1+x)^{-3}=e^{-10} \cdot 1=e^{-10}
\end{gathered}
$$

4) Assuming $\operatorname{tg} x=1+\alpha$, we get $\alpha \rightarrow 0$ when $x \rightarrow \frac{\pi}{4}$, and

$$
\begin{gathered}
\operatorname{tg} 2 x=\frac{2 \operatorname{tg} x}{1-\lg ^{2} x}=-\frac{2(\alpha+1)}{\alpha(\alpha+2)} \\
\lim _{x \rightarrow \frac{\pi}{4}}(\operatorname{tg} x)^{\operatorname{tg} 2 x}=\lim _{\alpha \rightarrow 0}\left[(1+\alpha)^{\frac{1}{\alpha}}\right]^{-\frac{2(a+1)}{\alpha+2}}=e^{-1}
\end{gathered}
$$

since

$$
\begin{aligned}
& \lim _{a \rightarrow 0} \frac{2(a+1)}{a+2}=1 \\
& \quad-44-
\end{aligned}
$$