Example. Compute the limit

$$
\lim _{x \rightarrow 3} \frac{x^{3}-4 x^{2}-3 x+18}{x^{3}-5 x^{2}+3 x+9}
$$
Solution.

1. The expression under the limit sign (a rational fraction) is the ratio of two infinitesimal functions as $x \rightarrow 3$.

Factorize the numerator and the denominator:

$$
\frac{x^{3}-4 x^{2}-3 x+18}{x^{3}-5 x^{2}+3 x+9}=\frac{(x-3)^{2}(x+2)}{(x-3)^{2}(x+1)}
$$

2. Since in the definition of the limit of a function as $x \rightarrow 3$, the argument cannot take the value 3, we can cancel the factor $(x-3)^{2}$. We get

$$
\lim _{x \rightarrow 3} \frac{x^{3}-4 x^{2}-3 x+18}{x^{3}-5 x^{2}+3 x+9}=\lim _{x \rightarrow 3} \frac{x+2}{x+1}=\frac{5}{4}
$$

Answer. $\lim _{x \rightarrow 3} \frac{x^{3}-4 x^{2}-3 x+18}{x^{3}-5 x^{2}+3 x+9}=\frac{5}{4}$.

Conditions of the Problems. Calculate the limit.

1. $\lim _{x \rightarrow-1} \frac{x^{3}+3 x^{2}+7 x+5}{x^{2}-x-2}$.
2. $\lim _{x \rightarrow 1} \frac{x^{2}-2 x+1}{2 x^{2}+x-3}$.
3. $\lim _{x \rightarrow-1} \frac{x^{3}+4 x^{2}+5 x+2}{x^{2}+2 x+1}$.
4. $\lim _{x \rightarrow 2} \frac{x^{3}-6 x^{2}+12 x-8}{x^{3}-2 x^{2}+2 x-4}$.
5. $\lim _{x \rightarrow 1} \frac{x^{3}-1}{2 x^{2}-x-1}$.
6. $\lim _{x \rightarrow-1} \frac{x^{3}+x^{2}-x-1}{x^{3}-3 x-2}$.
7. $\lim _{x \rightarrow 1} \frac{x^{4}-1}{x^{4}-x^{3}+x-1}$.
8. $\lim _{x \rightarrow-3} \frac{x^{2}+2 x-3}{x^{3}+5 x^{2}+6 x}$.
9. $\lim _{x \rightarrow-1} \frac{x^{4}+x^{3}+2 x+2}{x^{2}-1}$.
10. $\lim _{x \rightarrow 2} \frac{x^{3}-x^{2}-x-2}{x^{3}-2 x^{2}+x-2}$.

Answers. 1. 0. 2. 2/3. 3. 0. 4. 2. 5. 1. 6. $-4 / 3$. 7. 0. 8. $-1 / 2$. 9. $1.10 .7 / 3$.

## 3.8. Calculation of $\lim _{x \rightarrow 0}[f(x) / g(x)]$

Problem Statement. Calculate the limit

$$
\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}
$$

where $f(x)$ and $g(x)$ are infinitesimal functions at the point $x=0$.

Solution Plan. The infinitesimal functions in the numerator and the denominator can be replaced by their equivalent (tabular) functions.

If $f(x), f_{1}(x), g(x), g_{1}(x)$ are infinitesimal functions at the point $x=0$ such that $f(x) \sim f_{1}(x)$ and $g(x) \sim g_{1}(x)$ at the point $x=0$, and $\lim _{x \rightarrow 0} f_{1}(x) / g_{1}(x)$ exists, then $\lim _{x \rightarrow 0} f(x) / g(x)$ exists, and

$$
\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim _{x \rightarrow 0} \frac{f_{1}(x)}{g_{1}(x)}
$$