## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{3^{2 x}-7^{x}}{\arcsin 3 x-5 x}$
## Solution

$\lim _{x \rightarrow 0} \frac{3^{2 x}-7^{x}}{\arcsin 3 x-5 x}=\lim _{x \rightarrow 0} \frac{\left(9^{x}-1\right)-\left(7^{x}-1\right)}{\arcsin 3 x-5 x}=$

$=\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 9}\right)^{x}-1\right)-\left(\left(e^{\ln 7}\right)^{x}-1\right)}{\arcsin 3 x-5 x}=$

$=\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 9}-1\right)-\left(e^{x \ln 7}-1\right)}{\arcsin 3 x-5 x}=$

$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 9}-1\right)-\left(e^{x \ln 7}-1\right)\right)}{\frac{1}{x}(\arcsin 3 x-5 x)}=$

$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 9}-1\right)-\left(e^{x \ln 7}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(\arcsin 3 x-5 x)}=$

$=\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 9}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{x \ln 7}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\arcsin 3 x}{x}-\lim _{x \rightarrow 0} \frac{5 x}{x}\right)=$

Using the substitution of equivalent infinitesimals:

$e^{x \ln 9}-1 \sim x \ln 9$, as $x \rightarrow 0(x \ln 9 \rightarrow 0)$

$e^{x \ln 7}-1 \sim x \ln 7$, as $x \rightarrow 0(x \ln 7 \rightarrow 0)$

$\arcsin 3 x \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$

We get:

$=\frac{\lim _{x \rightarrow 0} \frac{x \ln 9}{x}-\lim _{x \rightarrow 0} \frac{x \ln 7}{x}}{\lim _{x \rightarrow 0} \frac{3 x}{x}-\lim _{x \rightarrow 0} 5}=\frac{\lim _{x \rightarrow 0} \ln 9-\lim _{x \rightarrow 0} \ln 7}{\lim _{x \rightarrow 0} 3-5}=$
$=\frac{\ln 9-\ln 7}{3-5}=\frac{\ln 7-\ln 9}{2}=\frac{1}{2} \ln \frac{7}{9}=\ln \sqrt{\frac{7}{9}}$

## Problem Kuznetsov Limits 15-19