518. Find the limits:
1) $\lim _{x \rightarrow 0} \frac{x}{\sin x}$; 2) $\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 5 x}$; 3) $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}$.
Solution. 1) $\lim _{x \rightarrow 0} \frac{x}{\sin x}=\lim _{x \rightarrow 0} \frac{1}{\frac{\sin x}{x}}=\frac{1}{1}=1$.

2) Transform the given expression so that the problem is reduced to limit (1):

$$
\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 5 x}=\lim _{x \rightarrow 0} \frac{\frac{\sin 4 x}{4 x} \cdot 4 x}{\frac{\sin 5 x}{5 x} \cdot 5 x}=\frac{4}{5} \lim _{x \rightarrow 0} \frac{\frac{\sin 4 x}{4 x}}{\frac{\sin 5 x}{5 x}}=\frac{4}{5}
$$

3) $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{x}{2}}{x^{2}}=\lim _{x \rightarrow 0} \frac{2 \cdot \sin \frac{x}{2} \cdot \sin \frac{x}{2}}{4 \cdot \frac{x}{2} \cdot \frac{x}{2}}=\frac{1}{2}$.

Find the limits: