54. 55) $\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}$; 2) $\lim _{x \rightarrow 0} \frac{x^{2}}{1-\cos x}$;
3) $\lim _{x \rightarrow 1} \frac{\cos \frac{\pi x}{2}}{1-x}$; 4) $\lim _{x \rightarrow-2} \frac{x^{2}-4}{\operatorname{arctg}(x+2)}$.
Solution. We establish that the given function is not defined at the limit point, and that with the given change in the argument, it represents the ratio of two infinitesimally small quantities (the case $\frac{0}{0}$). After this, we subject the function to transformations in order to use the 1st remarkable limit:

$$
\begin{aligned}
& \lim _{a \rightarrow 0} \frac{\sin \alpha}{\alpha}=1(\alpha-\text { radian measure of the angle) } \\
& \text { 1) } \lim _{x \rightarrow 0} \frac{\sin 3 x}{x}=\lim _{\frac{3}{\sin 3 x}}^{3 x}=3 \lim _{3 x \rightarrow 0} \frac{\sin 3 x}{3 x}=3 \cdot 1=3
\end{aligned}
$$

2) We apply the trigonometric formula $1-\cos x=$ $=2 \sin ^{2} \frac{x}{2}$ :

$$
\lim _{x \rightarrow 0} \frac{x^{2}}{1-\cos x}=\lim \frac{x^{2}}{2 \sin ^{2} \frac{x}{2}}=2\left(\lim \frac{\frac{x}{2}}{\sin \frac{x}{2}}\right)^{2}=2 \cdot 1=2
$$

3) Here, to use the 1st remarkable limit, we make a change of variable: $1-x=t$. Then when $x \rightarrow 1$, $t \rightarrow 0$ and

$$
\begin{gathered}
\lim _{x \rightarrow 1} \frac{\cos \frac{\pi x}{2}}{1-x}=\lim _{t \rightarrow 0} \frac{\cos \left(\frac{\pi}{2}-\frac{\pi}{2} t\right)}{t}=\lim \frac{\sin \frac{\pi}{2} t}{t}= \\
=\frac{\pi}{2} \lim \frac{\sin \frac{\pi}{2} t}{\frac{\pi}{2} t}=\frac{\pi}{2} \cdot 1=\frac{\pi}{2} \\
-37-
\end{gathered}
$$

4) Assuming $\operatorname{arctg}(x+2)=v$, we get $x+2=\operatorname{tg} v, \quad v \rightarrow 0$, when $x \rightarrow-2$, and

$\lim _{x \rightarrow-2} \frac{x^{2}-4}{\operatorname{arctg}(x+2)}=\lim _{v \rightarrow 0} \frac{(\operatorname{tg} v-2)^{2}-4}{v}=\lim \frac{(\operatorname{tg} v-4) \operatorname{tg} v}{v}=$ $=\lim \frac{\operatorname{tg} v-4}{\cos v} \cdot \operatorname{lnm} \frac{\sin v}{v}=-4 \cdot 1=-4$.