Example. Calculate the limit

$$
\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\tan^{2} 2 x}
$$
Solution.

1. Since

$$
\lim _{x \rightarrow \pi}[\cos 3 x-\cos x]=0, \quad \lim _{x \rightarrow \pi} \operatorname{tg}^{2} 2 x=0
$$

the expression under the limit sign is a ratio of two infinitesimal functions as $x \rightarrow \pi$. We need to replace these infinitesimal functions with equivalent ones. For this, we first make the substitution $x-\pi=t$:

$$
\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\operatorname{tg}^{2} 2 x}=\lim _{t \rightarrow 0} \frac{\cos 3(\pi+t)-\cos (\pi+t)}{\operatorname{tg}^{2} 2(\pi+t)}
$$

2. Using trigonometric formulas and replacing the product and quotient of infinitesimal functions with equivalent ones, we get

$$
\begin{aligned}
& \lim _{t \rightarrow 0} \frac{\cos 3(\pi+t)-\cos (\pi+t)}{\operatorname{tg}^{2} 2(\pi+t)}= \\
& \quad=\lim _{t \rightarrow 0} \frac{\cos t-\cos 3 t}{\operatorname{tg}^{2} 2 t}=\lim _{t \rightarrow 0} \frac{-2 \sin 2 t \sin (-t)}{\operatorname{tg}^{2} 2 t}=\lim _{t \rightarrow 0} \frac{2 \cdot 2 t \cdot t}{4 t^{2}}=1
\end{aligned}
$$

Answer. $\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\operatorname{tg}^{2} 2 x}=1$.

Conditions of the Problems. Calculate the limits.

1. $\lim _{x \rightarrow 1} \frac{x^{3}-1}{\ln x}$.
2. $\lim _{x \rightarrow \pi} \frac{1+\cos 5 x}{\sin ^{2} 3 x}$.
3. $\lim _{x \rightarrow 1 / 2} \frac{1+\cos 2 \pi x}{\operatorname{tg}^{2} 2 \pi x}$.
4. $\lim _{x \rightarrow 2} \frac{\sin 3 \pi x}{\sin 8 \pi x}$.
5. $\lim _{x \rightarrow 2} \frac{\sqrt{x^{2}-x-1}-1}{\ln (x-1)}$.
6. $\lim _{x \rightarrow \pi / 2} \frac{\operatorname{tg} 5 x}{\operatorname{tg} 3 x}$.
7. $\lim _{x \rightarrow 1} \frac{1-x^{3}}{\sin \pi x}$.
8. $\lim _{x \rightarrow 1} \frac{2-\sqrt{5-x}}{\sin \pi x}$.
9. $\lim _{x \rightarrow \pi} \frac{\operatorname{tg} 5 x}{\sin 3 x}$.
10. $\lim _{x \rightarrow 2} \frac{2^{x}-4}{\sin \pi x}$.

Answers. 1. 3. 2. 5/18. 3. $1 / 2$. 4. 3/8. 5. 3/2. 6. 3/5. 7. $3 / \pi$. 8. $-1 /(4 \pi)$. 9. $-5 / 3$. 10. $(4 \ln 2) / \pi$.

### 3.10. Calculation of $\lim _{x \rightarrow 0}\left[u(x)^{v(x)}\right]$

Problem Statement. Calculate the limit

$$
\lim _{x \rightarrow 0}\left[u(x)^{v(x)}\right]
$$

where $\lim _{x \rightarrow 0} u(x)=1$ and $\lim _{x \rightarrow 0} v(x)=\infty$.

## PLAN OF SOLUTION.

1. Transform the expression under the limit sign:

$$
u(x)^{v(x)}=e^{v(x) \ln u(x)}
$$

2. Since the exponential function $e^{x}$ is continuous, we can take the limit under the sign of this function. We have

$$
\lim _{x \rightarrow 0}\left[u(x)^{v(x)}\right]=\lim _{x \rightarrow 0} e^{v(x) \ln u(x)}=e^{\lim _{x \rightarrow 0}[v(x) \ln u(x)]}
$$

3. Calculate the limit of the exponent

$$
\lim _{x \rightarrow 0}[v(x) \ln u(x)]
$$

replacing infinitesimal functions with equivalent ones.

4. Write the final answer.