## Example. Compute the limit

$$
\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}
$$
Solution. The expression under the limit sign is the ratio of two infinitesimals at the point $x=0$, since

$$
\lim _{x \rightarrow 0}(2 x \sin x)=0, \quad \lim _{x \rightarrow 0}(1-\cos x)=0
$$

The infinitesimals in the numerator and denominator are replaced by equivalent ones:

$$
\begin{array}{ll}
2 x \sin x \sim 2 x \cdot x, & x \rightarrow 0 \\
1-\cos x \sim \frac{x^{2}}{2}, & x \rightarrow 0
\end{array}
$$

Thus,

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

Answer. $\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}=4$.

Conditions of the Problems. Calculate the limit.

1. $\lim _{x \rightarrow 0} \frac{\ln (1+\sin 2 x)}{\sin 3 x}$.
2. $\lim _{x \rightarrow 0} \frac{3 x^{2}+6 x}{\sin 3 x}$.
3. $\lim _{x \rightarrow 0} \frac{5^{x}-1}{\ln (1+x)}$.
4. $\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{\cos 5 x-\cos 3 x}$.
5. $\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{e^{2 x^{2}}-1}$.
6. $\lim _{x \rightarrow 0} \frac{\sqrt{9+x}-3}{3 \operatorname{arctg} 2 x}$.
7. $\lim _{x \rightarrow 0} \frac{\operatorname{tg} 2 x}{e^{2 x}-1}$.
8. $\lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{\sin ^{2} x}$.
9. $\lim _{x \rightarrow 0} \frac{\sin 2 x}{\ln (1-2 x)}$.
10. $\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{\ln (e-2 x)-1}$.

Answers. 1. 2/3. 2. 2. 3. $\ln 5$.
4. $-1 / 4$.
5. 1 .
6. $1 / 12$. 7. 1. 8. $1 / 4$. 9. -1. 10. $-e$.

3.9. Calculation of $\lim _{x \rightarrow a}[f(x) / g(x)]$

PROBLEM STATEMENT. Calculate the limit

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

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

SOLUTION PLAN.

1. We need to replace $f(x)$ and $g(x)$ with equivalent infinitesimal functions. However, the table of equivalent infinitesimal functions is compiled for the point $x=0$. Therefore, we first make a variable substitution $x-a=t$ and will find the limit as $t \rightarrow 0$.
2. Transform the expression under the limit sign using algebraic and trigonometric formulas, and replace the infinitesimal functions in the product and quotient with equivalent ones.