Example 1. Find $\lim _{x \rightarrow 2}\left(4 x^{2}-6 x+3\right)$.
Solution. Since the limit of the algebraic sum of variables is equal to the same algebraic sum of the limits of these variables (formula (1.42)), a constant multiplier can be factored out of the limit sign (formula (1.44)), the limit of an integer positive power is equal to the same power of the limit (formula (1.45)), and the limit of a constant is the constant itself (formula (1.48)), we sequentially obtain

$$
\begin{gathered}
\lim _{x \rightarrow 2}\left(4 x^{2}-6 x+3\right)=\lim _{x \rightarrow 2} 4 x^{2}-\lim _{x \rightarrow 2} 6 x+\lim _{x \rightarrow 2} 3= \\
=4 \lim _{x \rightarrow 2} x^{2}-6 \lim _{x \rightarrow 2} x+\lim _{x \rightarrow 2} 3=4\left(\lim _{x \rightarrow 2} x\right)^{2}-6 \lim _{x \rightarrow 2} x+3= \\
=4 \cdot 2^{2}-6 \cdot 2+3=16-12+3=7 .
\end{gathered}
$$

Remark 1. The calculation of the limit of a second-degree polynomial has been reduced to calculating its value at the limit of the argument.

Remark 2. To calculate the limit of an n-th degree polynomial

$$
P_{n}(x)=b_{0}+b_{1} x+b_{2} x^{2}+\ldots+b_{n} x^{n} \text { as } x \rightarrow b
$$

it is sufficient to find \( P_{n}(b) \), i.e., its value at \( x=b \). For example,

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

Example 2. Find \(\lim _{x \rightarrow 1} \frac{3 x^{2}-4 x+7}{2 x^{2}-5 x+6}\).

Solution. Since the limit of a quotient is equal to the quotient of the limits, based on formula (1.46), and using formulas (1.42), (1.44), (1.45), and (1.48), we sequentially obtain

$$
\lim _{x \rightarrow 1} \frac{3 x^{2}-4 x+7}{2 x^{2}-5 x+6}=\frac{\lim _{x \rightarrow 1}\left(3 x^{2}-4 x+7\right)}{\lim _{x \rightarrow 1}\left(2 x^{2}-5 x+6\right)}=\frac{\lim _{x \rightarrow 1} 3 x^{2}-\lim _{x \rightarrow 1} 4 x+\lim _{x \rightarrow 1} 7}{\lim _{x \rightarrow 1} 2 x^{2}-\lim _{x \rightarrow 1} 5 x+\lim _{x \rightarrow 1} 6}=
$$

$$
=\frac{3 \lim _{x \rightarrow 1} x^{2}-4 \lim _{x \rightarrow 1} x+\lim _{x \rightarrow 1} 7}{2 \lim _{x \rightarrow 1} x^{2}-5 \lim _{x \rightarrow 1} x+\lim _{x \rightarrow 1} 6}=\frac{3 \cdot 1^{2}-4 \cdot 1+7}{2 \cdot 1^{2}-5 \cdot 1+6}=\frac{6}{3}=2
$$

Remark 1. The calculation of the limit of a rational function (the ratio of two polynomials) has been reduced to calculating the value of this function at the limit of the argument.

Remark 2. To calculate the limit of a rational function

$$
R(x)=\frac{b_{0}+b_{1} x+b_{2} x^{2}+\ldots+b_{n} x^{n}}{c_{0}+c_{1} x+c_{2} x^{2}+\ldots+c_{m} x^{m}}
$$

as \( x \rightarrow a \), it is sufficient to find its value at \( x=a \).

For example,

$$
\lim _{x \rightarrow 2} \frac{x^{5}-2 x^{4}+3 x^{2}-6 x+9}{3 x^{4}-12 x^{2}+7 x-11}=\frac{2^{5}-2 \cdot 2^{4}+3 \cdot 2^{2}-6 \cdot 2+9}{3 \cdot 2^{4}-12 \cdot 2^{2}+7 \cdot 2-11}=\frac{9}{3}=3
$$

Remark 3. The limit of an elementary function \( f(x) \) as \( x \rightarrow a \), where \( a \) belongs to its domain, is equal to the value of the function at \( x=a \), i.e.,

$$
\lim _{x \rightarrow a} f(x)=f(a)
$$

For example,

$$
\text { 1) } \begin{gathered}
\lim _{x \rightarrow 4} \frac{\sqrt{x^{2}-7}+\sqrt[3]{x+60}}{2 x^{2}-5 x-8}=\frac{\sqrt{4^{2}-7}+\sqrt[3]{4+60}}{2 \cdot 4^{2}-5 \cdot 4-8}= \\
=\frac{\sqrt{9}+\sqrt[3]{64}}{32-20-8}=\frac{3+4}{4}=\frac{7}{4}
\end{gathered}
$$

2). \(\lim \left[\lg \left(t+\sqrt{t^{2}+80}\right)\right]+t \sqrt{t^{2}+8}=\lg (1+\sqrt{81})+1 \sqrt{1+8}=1+3=4\). \(t \rightarrow 1\)