71. 72) $\lim _{x \rightarrow \infty} \frac{3 x^{2}-1}{5 x^{2}+2 x}$
2) $\lim _{n \rightarrow-\infty} \frac{n}{\sqrt{n^{2}+1}}$
3) $\lim _{n \rightarrow+\infty} \frac{1+7^{n+2}}{3-7^{n}}$
4) $\lim _{n \rightarrow+\infty} \frac{2+4+6+\ldots+2 n}{1+3+5+\ldots+(2 n+1)}$
5) $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan 2 x}{\cot\left(\frac{\pi}{4}-x\right)}$

$6)^{*} \lim _{n \rightarrow+\infty} \frac{n^{3}}{n^{2}+2^{2}+3^{2}+\ldots+n^{2}}$.
Solution. Having made sure that the case is $\frac{\infty}{\infty}$, we subject the function to transformations.

1) Dividing the numerator and denominator of the fraction by $x^{2}$ (the highest power of $x$ here), we find

$$
\lim _{x \rightarrow \infty} \frac{3 x^{2}-1}{5 x^{2}+2 x}=\lim \frac{3-\frac{1}{x^{2}}}{5+\frac{2}{x}}=\frac{3-0}{5+0}=\frac{3}{5}
$$

since when $x \rightarrow \infty$, the quantities $\frac{1}{x^{2}}$ and $\frac{1}{x}$ are infinitesimals. This problem can also be solved by a change of variable. Setting $x=\frac{1}{\alpha}$, we get: $\alpha \rightarrow 0$ when $x \rightarrow \infty$ and

$$
\lim _{x \rightarrow \infty} \frac{3 x^{2}-1}{5 x^{2}+2 x}=\lim _{a \rightarrow 0} \frac{\frac{3}{\alpha^{2}}-1}{\frac{5}{\alpha^{2}}+\frac{2}{a}}=\lim \frac{3-a^{2}}{5+2 a}=\frac{3}{5}
$$

In general, the limit transition as $x \rightarrow \infty$ can always be replaced by a limit transition as $\alpha \rightarrow 0$, if the new independent variable is taken to be the reciprocal of the original variable, i.e., if we set $\alpha=\frac{1}{x}$.

2) This problem can be solved in the same two ways as the previous one.

Dividing the numerator and denominator by $n$, we get

$$
\lim _{n \rightarrow-\infty} \frac{n}{\sqrt{n^{2}+1}}=\operatorname{lin} \frac{1}{-\sqrt{1+\frac{1}{n^{2}}}}=\frac{1}{-1}=-1
$$

or, setting $n=\frac{1}{\alpha}$, we find $\alpha \rightarrow-0$ when $n \rightarrow-\infty$, and

$$
\lim _{n \rightarrow-\infty} \frac{n}{\sqrt{n^{2}+1}}=\lim _{a \rightarrow-0} \frac{\frac{1}{a}}{\sqrt{\frac{1}{a^{2}}+1}}=\lim \frac{1}{-\sqrt{1+a^{2}}}=-1
$$

Here the minus sign appears due to the introduction under the square root sign (in the first solution) or the extraction from this sign (in the second solution) of a negative divisor, and if $a<0$, then

$$
a \sqrt{\bar{b}}=-\sqrt{a^{2} b} \text { and } \sqrt{a^{2} b}=-a \sqrt{\bar{b}}
$$

From this solution, it follows that when $n \rightarrow+\infty$, the limit of the given function will be one, and when $n \rightarrow \infty$, the limit of this function does not exist.

3) Multiplying the numerator and denominator of the fraction by $7^{-n}$, we get

$$
\lim _{n \rightarrow+\infty} \frac{1+7^{n+2}}{3-7^{n}}=\lim \frac{7^{-n}+7^{2}}{3 \cdot 7^{-n}-1}=\frac{0+49}{0-1}=-49
$$

since $\lim _{n \rightarrow+\infty} 7^{-n}=7^{-\infty}=0$.

4) Here the numerator of the fraction is the sum of $n$ terms of an arithmetic progression, and the denominator is the sum of $n+1$ terms of another arithmetic progression. Transforming them using the known formula, we get

$$
\begin{aligned}
& \lim _{n \rightarrow+\infty} \frac{2+4+6+\ldots+2 n}{1+3+5+\ldots+(2 n+1)}= \\
& =\lim \frac{\frac{2+2 n}{2} n}{\frac{1+2 n+1}{2}(n+1)}=\lim \frac{n}{n+1}=\lim \frac{1}{1+\frac{1}{n}}=1
\end{aligned}
$$

5) Identically transforming the fraction so that we can then cancel it by a factor tending to zero:

$$
\begin{gathered}
\lim _{x \rightarrow \frac{\pi}{4}} \frac{\operatorname{tg} 2 x}{\operatorname{ctg}\left(x-\frac{\pi}{4}\right)}=\lim \frac{\sin 2 x \sin \left(x-\frac{\pi}{4}\right)}{\cos 2 x \cos \left(x-\frac{\pi}{4}\right)}= \\
=\lim \frac{\sin 2 x}{\cos \left(x-\frac{\pi}{4}\right)} \lim \frac{\sin \left(x-\frac{\pi}{4}\right)}{\cos 2 x}= \\
=\frac{1}{1} \cdot \operatorname{li} n \frac{\sin \left(x-\frac{\pi}{4}\right)}{\sin \left(\frac{\pi}{2}-2 x\right)}=\lim \frac{-\sin \left(\frac{\pi}{4}-x\right)}{2 \sin \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-x\right)}=-\frac{1}{2}
\end{gathered}
$$

6)* Transforming the denominator using the formula for the sum of squares of a natural number series:

$$
1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}
$$

we get

$$
\begin{gathered}
\lim _{n \rightarrow+\infty} \frac{n^{3}}{1^{2}+2^{2}+3^{2}+\ldots+n^{2}}=\lim \frac{6 n^{3}}{n(n+1)(2 n+1)}= \\
=\lim \frac{6}{\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}=3
\end{gathered}
$$