## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0}(1-\ln (\cos x))^{\frac{1}{\operatorname{tg}^{2} x}}
$$
## Solution

$\lim _{x \rightarrow 0}(1-\ln (\cos x))^{\frac{1}{\operatorname{tg}^{2} x}}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln (1-\ln (\cos x))}\right)^{\frac{1}{\operatorname{tg}^{2} x}}=$
$=\lim _{x \rightarrow 0} e^{\ln (1-\ln (\cos x)) / \operatorname{tg}^{2} x}=$
$=\exp \left\{\lim _{x \rightarrow 0} \frac{\ln (1-\ln (\cos x))}{\operatorname{tg}^{2} x}\right\}=$

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_5defd88eaa3a4fc3f868g-12.jpg?height=86&width=1350&top_left_y=453&top_left_x=152)
$\operatorname{tg} x \sim x$, as $x \rightarrow 0$

We get:

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1-2 \sin ^{2} \frac{x}{2}\right) \sim-2 \sin ^{2} \frac{x}{2}, \quad$ as $\quad x \rightarrow 0\left(-2 \sin ^{2} \frac{x}{2} \rightarrow 0\right)$

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{x}{2}}{x^{2}}\right\}=$

Using the substitution of equivalent infinitesimals:

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

We get:

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

$=\exp \left\{\frac{1}{2}\right\}=e^{\frac{1}{2}}$

## Problem Kuznetsov Limits 17-29