## Problem Statement

Calculate the limit of the function:

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

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

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

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

Using the substitution of equivalent infinitesimals:

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

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

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{-\sin ^{2} \frac{x}{2}}{\operatorname{tg}^{2} 3 x}\right\}=$

Using the substitution of equivalent infinitesimals:

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

$\operatorname{tg} 3 x \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{-\left(\frac{x}{2}\right)^{2}}{(3 x)^{2}}\right\}=\exp \left\{\lim _{x \rightarrow 0} \frac{-\left(\frac{1}{4}\right)}{9}\right\}=e^{-\frac{1}{36}}$

Problem Kuznetsov Limits 17-30