## Problem Statement

Calculate the limit of the function:

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

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

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

$=\lim _{x \rightarrow 0} e^{3 \cdot \ln \left(1-\ln \left(1+x^{3}\right)\right) /\left(x^{2} \arcsin x\right)}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{3 \cdot \ln \left(1-\ln \left(1+x^{3}\right)\right)}{x^{2} \arcsin x}\right\}=$

Using the substitution of equivalent infinitesimals:

$\ln \left(1-\ln \left(1+x^{3}\right)\right) \sim-\ln \left(1+x^{3}\right)_{\text {, as }} x \rightarrow 0\left(-\ln \left(1+x^{3}\right) \rightarrow 0\right)$ $\arcsin x \sim x$, as $x \rightarrow 0$

We get:

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1+x^{3}\right) \sim x_{, \text {as }}^{3} x \rightarrow 0\left(x^{3} \rightarrow 0\right)$

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{-3 \cdot x^{3}}{x^{3}}\right\}=\exp \left\{\lim _{x \rightarrow 0} \frac{-3}{1}\right\}=$

$=\exp \left\{\frac{-3}{1}\right\}=\exp \{-3\}=e^{-3}$

## Problem Kuznetsov Limits 17-1