## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0}\left(2-e^{x^{2}}\right)^{1 / \ln \left(1+\operatorname{tg}^{2}\left(\frac{\pi x}{3}\right)\right)}$
## Solution

$\lim _{x \rightarrow 0}\left(2-e^{x^{2}}\right)^{1 / \ln \left(1+\operatorname{tg}^{2}\left(\frac{\pi x}{3}\right)\right)}=$

$=\lim _{x \rightarrow 0}\left(e^{\ln \left(2-e^{x^{2}}\right)}\right)^{1 / \ln \left(1+\operatorname{tg}^{2}\left(\frac{\pi x}{3}\right)\right)}=$

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1-\left(e^{x^{2}}-1\right)\right) \sim-\left(e^{x^{2}}-1\right)$, as $x \rightarrow 0\left(-\left(e^{x^{2}}-1\right) \rightarrow 0\right)$ $\ln \left(1+\operatorname{tg}^{2}\left(\frac{\pi x}{3}\right)\right) \sim \operatorname{tg}^{2}\left(\frac{\pi x}{3}\right)$, as $x \rightarrow 0\left(\operatorname{tg}^{2}\left(\frac{\pi x}{3}\right) \rightarrow 0\right)$

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{-\left(e^{x^{2}}-1\right)}{\operatorname{tg}^{2}\left(\frac{\pi x}{3}\right)}\right\}=$

Using the substitution of equivalent infinitesimals:

$e^{x^{2}}-1 \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$

![](https://cdn.mathpix.com/cropped/2024_05_22_c071f2ede8d7a98b06e8g-12.jpg?height=136&width=827&top_left_y=1626&top_left_x=150)

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{x \rightarrow 0} \frac{-x^{2}}{\left(\frac{\pi x}{3}\right)^{2}}\right\}=\exp \left\{\lim _{x \rightarrow 0} \frac{-x^{2}}{\frac{\pi^{2} x^{2}}{9}}\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0} \frac{-9}{\pi^{2}}\right\}=\exp \left\{-\frac{9}{\pi^{2}}\right\}=e^{-\frac{9}{\pi^{2}}}
\end{aligned}
$$

## Problem Kuznetsov Limits 17-17