Example 14. Find $\lim _{n \rightarrow \infty} \frac{12 n+5}{\sqrt[3]{27 n^{3}+6 n^{2}+8}}$.
Solution. As $n \rightarrow \infty$, the numerator and denominator also tend to infinity, resulting in an indeterminate form $\frac{\infty}{\infty}$. To find the limit, we divide the numerator and denominator by $\boldsymbol{n}$ and bring $\boldsymbol{n}$ under the root sign:

$$
\lim _{n \rightarrow \infty} \frac{12 n+5}{\sqrt[3]{27 n^{3}+6 n^{2}+8}}=\lim _{n \rightarrow \infty} \frac{12+\frac{5}{n}}{\sqrt[3]{27+\frac{6}{n}+\frac{8}{n^{3}}}}=\frac{12}{\sqrt[3]{27}}=\frac{12}{3}=4
$$

(Here it is taken into account that $\lim _{n \rightarrow \infty} \frac{5}{n}=0, \lim _{n \rightarrow \infty} \frac{6}{n}=0, \lim _{n \rightarrow \infty} \frac{8}{n^{3}}=0$.)

## Problems

Find the limits:

1. $\lim _{x \rightarrow 3}\left(2 x^{2}-7 x+6\right)$.
2. $\lim _{x \rightarrow 1}\left(3 x^{4}-5 x^{3}+6 x^{2}-4 x+7\right)$.
3. $\lim _{x \rightarrow 2} \frac{4 x^{2}-5 x+2}{3 x^{2}-6 x+4}$.
4. $\lim _{x \rightarrow 4} \frac{x^{2}-5 x+4}{x^{2}-7 x+6}$.
5. $\lim _{x \rightarrow 5} \frac{x^{2}-7 x+12}{x^{2}-6 x+5}$.
6. $\lim _{t \rightarrow 3}\left[2 t \sqrt{t^{2}-8}+\lg \left(3 t+\sqrt{t^{2}-8}\right)-\frac{\sqrt[3]{t^{2}-1}+\sqrt[4]{t^{3}-2 t^{2}+7}}{t^{3}-2 t^{2}+t-11}\right]$.
7. $\lim _{x \rightarrow 6} \frac{x^{2}-8 x+12}{x^{2}-7 x+6}$.
8. $\lim _{x \rightarrow 2} \frac{3 x^{2}-7 x+2}{4 x^{2}-5 x-6}$.
9. $\lim _{x \rightarrow 1} \frac{x^{3}-3 x+2}{x^{4}-4 x+3}$.
10. $\lim _{x \rightarrow \infty} \frac{10 x^{3}-6 x^{2}+7 x+5}{8-4 x+3 x^{2}-2 x^{3}}$.
11. $\lim _{x \rightarrow \infty} \frac{2 x^{4}-5 x^{3}+7 x^{2}+8 x-9}{3 x^{5}-6 x^{3}+4 x^{2}-2 x+11}$.
12. $\lim _{x \rightarrow \infty} \frac{x^{7}+8 x^{6}+5 x^{4}-3 x^{2}-12}{10 x^{6}+7 x^{5}-6 x^{3}-4 x+17}$.
13. $\lim _{x \rightarrow 5} \frac{x^{2}-25}{2-\sqrt{x-1}}$.
14. $\lim _{n \rightarrow \infty} \frac{\sqrt{4 n^{2}-3}}{2 n+1}$.
15. $\lim _{x \rightarrow \infty} \frac{6 x-5}{1+\sqrt{x^{2}+3}}$.
16. $\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^{2}-1}$.
17. $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin 2 x-\cos 2 x-1}{\cos x-\sin x}$.
18. $\lim _{x \rightarrow \pi} \frac{\sqrt{1-\operatorname{tg} x}-\sqrt{1+\operatorname{tg} x}}{\sin 2 x}$.
19. $\lim _{x \rightarrow \infty}\left(\frac{2 x^{2}}{1-x^{2}}+3^{\frac{1}{x}}\right)$.

## Answers

1.3. 2. 7. 3. 2. 4.0. 5. $\infty$. 6.4. 7. $\frac{4}{5}$. 8. $\frac{15}{44}$. 9. $\frac{1}{2}$ 10. -5 . 11. 0 . 12. $\infty$.

13. -40 . 14. $\frac{3}{4}$. Hint. Let $x=t^{12}$. 15. 1. 16.6. 17. $\frac{1}{2}$.
14. $-\sqrt{2} \cdot 19 \cdot-\frac{1}{2} \cdot 20 \cdot-1$.

§ 1.5. The number e, $\lim _{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha}=1$

1. The number e is the limit

$$
\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e
$$

or

$$
\lim _{\alpha \rightarrow 0}(1+\alpha)^{\frac{1}{\alpha}}=e
$$

The number $e$ is useful for resolving indeterminate forms of the type $1^{\infty}$.

If the base of logarithms is the number $e$, then the logarithms are called natural:

$$
\ln x=\log _{e} x
$$

2. If the angle $\alpha$ is expressed in radians, then

$$
\lim _{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha}=1
$$

3. When finding limits of the form

$$
\lim _{x \rightarrow a}[\varphi(x)]^{\psi(x)}=C
$$

the following should be kept in mind:

1) if there exist finite limits

$$
\lim _{x \rightarrow a} \varphi(x)=A \text { and } \lim _{x \rightarrow a} \psi(x)=B
$$

then

$$
C=A^{B}
$$

2) if

$$
\lim _{x \rightarrow a} \varphi(x)=A \neq 1 \text { and } \lim _{x \rightarrow a} \psi(x)=\infty \text {, }
$$

then the limit is found using formulas (1.55) and (1.56);

3) if

$$
\lim _{x \rightarrow a} \varphi(x)=1, \lim _{x \rightarrow a} \psi(x)=\infty
$$

then we set $\varphi(x)=1+\alpha(x)$, where $\alpha(x) \rightarrow 0$ as $x \rightarrow a$ and, consequently,

$$
\begin{gathered}
C=\lim _{x \rightarrow a}\left\{[1+\alpha(x)]^{\frac{1}{\alpha(x)}}\right\}^{\alpha(x) \psi(x)}=e^{\lim _{x \rightarrow \alpha} \alpha(x) \psi(x)}= \\
=e^{\lim _{x \rightarrow a}[\varphi(x)-1] \psi(x)}
\end{gathered}
$$