497. Find the limits:
1) $\lim _{x \rightarrow 2}\left(3 x^{2}-2 x+7\right)$
2) $\lim _{x \rightarrow 3} \frac{3 x^{2}-1}{2 x^{3}+6 x^{2}-5}$;
3) $\lim _{x \rightarrow 2} \frac{x-2}{x^{2}+3 x-1}$;
4) $\lim _{x \rightarrow 3} \frac{x^{2}-2 x+1}{x-3}$;
5) $\lim _{x \rightarrow 0} \frac{5 x^{3}+2 x^{2}-x}{3 x}$;
6) $\lim _{x \rightarrow 4} \frac{x^{2}-6 x+8}{x^{2}-5 x+4}$;
7) $\lim _{x \rightarrow 1} \frac{\sqrt{1+8 x}-3}{\sqrt{4 x}-2}$;
8) $\lim _{x \rightarrow 0} x \sin \frac{1}{x}$;
9) $\lim _{x \rightarrow \infty} \frac{2 x^{3}-x+1}{5 x^{3}+2 x^{2}-1}$;
10) $\lim _{n \rightarrow \infty} \frac{\sqrt{n^{2}+1}+\sqrt{n}}{\sqrt[4]{n^{3}+n}-n}$.
Solution. 1) Applying the theorems on limits, we get $\lim _{x \rightarrow 2}\left(3 x^{2}-2 x+7\right)=3 \lim _{x \rightarrow 2} x \cdot \lim _{x \rightarrow 2} x-2 \lim _{x \rightarrow 2} x+\lim _{x \rightarrow 2} 7=3 \cdot 2 \cdot 2-$ $-2 \cdot 2+7=15$

2) The limits of the numerator and denominator exist and the limit of the denominator is not zero. Using the theorem on the limit of a quotient, we get

$$
\lim _{x \rightarrow 3} \frac{3 x^{2}-1}{2 x^{3}+6 x^{2}-5}=\frac{\lim _{x \rightarrow 3}\left(3 x^{2}-1\right)}{\lim _{x \rightarrow 3}\left(2 x^{3}+6 x^{2}-5\right)}=\frac{3 \cdot 9-1}{2 \cdot 27+6 \cdot 9-5}=\frac{26}{103}
$$

3) The limits of the numerator and denominator exist and the limit of the denominator is not zero.

Using the theorem on the limit of a quotient, we find that

$$
\lim _{x \rightarrow 2} \frac{x-2}{x^{2}+3 x-1}=\frac{0}{9}=0
$$

The obtained result can also be written as:

$$
\frac{x-2}{x^{2}+3 x-1} \rightarrow 0 \text { as } x \rightarrow 2
$$

Therefore, the quantity $\frac{x-2}{x^{2}+3 x-1}$ is infinitesimal as $x \rightarrow 2$.

4) We cannot directly apply the theorem on the limit of a quotient, as the limit of the denominator is zero (the denominator contains an infinitesimal quantity as $x \rightarrow 3$). In the numerator, we have a bounded quantity, different from zero. Thus, under the limit sign, we have the product of a bounded quantity $x^{2}-2 x+1$, different from zero, and an infinitely large quantity $\frac{1}{x-3}$ as $x \rightarrow 3$, which is the reciprocal of an infinitesimal quantity. Therefore,

$$
\frac{x^{2}-2 x+1}{x-3} \rightarrow \infty \text { as } x \rightarrow 3
$$

This can also be written as:

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

Therefore, the quantity $\frac{x^{2}-2 x+1}{x-3}$ is infinitely large as $x \rightarrow 3$.

5) In this example, we also cannot directly apply the theorem on the limit of a quotient, as the limits of the denominator and numerator are both zero, leading to an indeterminate form of $\frac{0}{0}$.

In such examples, when the numerator and denominator are polynomials, they must be factored, the fraction simplified, and then the limit taken.

Let's do this:

$\lim _{x \rightarrow 0} \frac{5 x^{3}+2 x^{2}-x}{3 x}=\lim _{x \rightarrow 0} \frac{x\left(5 x^{2}+2 x-1\right)}{3 x}=\lim _{x \rightarrow 0} \frac{5 x^{2}+2 x-1}{3}=-\frac{1}{3}$.

It should be noted that $x$ approaches zero but is not equal to zero; its limit is zero.

6) Here we also have an indeterminate form of $\frac{0}{0}$:

$\lim _{x \rightarrow 4} \frac{x^{2}-6 x+8}{x^{2}-5 x+4}=\lim _{x \rightarrow 4} \frac{(x-2)(x-4)}{(x-1)(x-4)}=\lim _{x \rightarrow 4} \frac{x-2}{x-1}=\frac{2}{3}$.

7) In this example, we have an indeterminate form of $\frac{0}{0}$.

If there is an irrationality under the limit sign, to resolve the indeterminate form of $\frac{0}{0}$, it is necessary to move the irrationality from the numerator to the denominator or from the denominator to the numerator, sometimes both, simplify the resulting fraction, and then take the limit

$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\sqrt{1+8 x}-3}{\sqrt{4 x}-2}=\lim _{x \rightarrow 1} \frac{(\sqrt{1+8 x}-3)(\sqrt{1+8 x}+3)(\sqrt{4 x}+2)}{(\sqrt{4 x}-2)(\sqrt{1+8 x}+3)(\sqrt{4 x}+2)}= \\
& =\lim _{x \rightarrow 1} \frac{(1+8 x-9)(\sqrt{4 x}+2)}{(\sqrt{1+8 x}+3)(4 x-4)}=\lim _{x \rightarrow 1} \frac{2(\sqrt{4 x}+2)}{\sqrt{1+8 x}+3}=\frac{4}{3}
\end{aligned}
$$

8) As $x \rightarrow 0$, the variable $x$ is an infinitesimal quantity, and $\left|\sin \frac{1}{x}\right| \leqslant 1$ for all $x \neq 0$. Therefore, the quantity $x \sin \frac{1}{x}$ - the product of an infinitesimal and a bounded quantity - will also be an infinitesimal quantity, so its limit $\lim _{x \rightarrow 0} x \sin \frac{1}{x}=0$.
9) The theorem on the limit of a quotient cannot be applied here, as the numerator and denominator do not have finite limits. In this case, we say: we have an indeterminate form of $\frac{\infty}{\infty}$. In such examples, to resolve the indeterminate form, it is advisable to divide the numerator and denominator by the power of $x$ with the highest exponent, and then take the limit

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

10) We have an indeterminate form of $\frac{\infty}{\infty}$. Dividing the numerator and denominator by $n$, we get
$\lim _{n \rightarrow \infty} \frac{\sqrt{n^{2}+1}+\sqrt{n}}{\sqrt[4]{n^{3}+n}-n}=\lim _{n \rightarrow \infty} \frac{\sqrt{1+\frac{1}{n^{2}}}+\sqrt{\frac{1}{n}}}{\sqrt[4]{\frac{1}{n}+\frac{1}{n^{3}}}-1}=\frac{1}{-1}=-1$.