## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0}\left(\frac{1+\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}\right)^{\frac{1}{\sin x^{3}}}$
## Solution

$\lim _{x \rightarrow 0}\left(\frac{1+\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}\right)^{\frac{1}{\sin x^{3}}}=$

$$
\begin{aligned}
& =\lim _{x \rightarrow 0}\left(e^{\ln \left(\frac{1+\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}\right)}\right)^{\frac{1}{\sin x^{3}}}= \\
& =\lim _{x \rightarrow 0} e^{\frac{1}{\sin x^{3}} \cdot \ln \left(\frac{1+\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}\right)}= \\
& =\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\sin x^{3}} \cdot \ln \left(\frac{1+\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}\right)\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\sin x^{3}} \cdot \ln \left(\frac{1+\sin x \cdot \cos 3 x-\sin x \cdot \cos 3 x+\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}\right)\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\sin x^{3}} \cdot \ln \left(1-\frac{\sin x \cdot \cos 3 x-\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}\right)\right\}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\ln \left(1-\frac{\sin x \cdot \cos 3 x-\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}\right) \sim-\frac{\sin x \cdot \cos 3 x-\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}$, as $x \rightarrow 0\left(-\frac{\sin x \cdot \cos 3 x-\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x} \rightarrow 0\right)$

$\sin x \sim x$, as $x \rightarrow 0$

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{x \rightarrow 0}-\frac{1}{x^{3}} \cdot \frac{\sin x \cdot \cos 3 x-\sin x \cdot \cos 2 x}{1+\sin x \cdot \cos 3 x}\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0}-\frac{\sin x}{x^{3}} \cdot \frac{\cos 3 x-\cos 2 x}{1+\sin x \cdot \cos 3 x}\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0}-\frac{\sin x}{x^{3}} \cdot \frac{-2 \sin \frac{3 x+2 x}{2} \sin \frac{3 x-2 x}{2}}{1+\sin x \cdot \cos 3 x}\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0}-\frac{\sin x}{x^{3}} \cdot \frac{-2 \sin \frac{5 x}{2} \sin \frac{x}{2}}{1+\sin x \cdot \cos 3 x}\right\}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\sin x \sim x$, as $x \rightarrow 0$

$\sin \frac{5 x}{2} \sim \frac{5 x}{2}$, as $x \rightarrow 0\left(\frac{5 x}{2} \rightarrow 0\right)$

![](https://cdn.mathpix.com/cropped/2024_05_22_b2fb00a8d2844ce741f0g-12.jpg?height=133&width=688&top_left_y=2183&top_left_x=163)

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{x \rightarrow 0}-\frac{x}{x^{3}} \cdot \frac{-2 \cdot \frac{5 x}{2} \cdot \frac{x}{2}}{1+\sin x \cdot \cos 3 x}\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0} \frac{2 \cdot \frac{5}{2} \cdot \frac{1}{2}}{1+\sin x \cdot \cos 3 x}\right\}=
\end{aligned}
$$

$$
\begin{aligned}
& =\exp \left\{\lim _{x \rightarrow 0} \frac{5}{2(1+\sin x \cdot \cos 3 x)}\right\}= \\
& =\exp \left\{\frac{5}{2(1+\sin 0 \cdot \cos (3 \cdot 0))}\right\}= \\
& =\exp \left\{\frac{5}{2}\right\}=e^{\frac{5}{2}}
\end{aligned}
$$

## Problem Kuznetsov Limits 17-23