## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{10^{2 x}-7^{-x}}{2 \tan x-\arctan x}$
## Solution

$\lim _{x \rightarrow 0} \frac{10^{2 x}-7^{-x}}{2 \tan x-\arctan x}=\lim _{x \rightarrow 0} \frac{\left(100^{x}-1\right)-\left(7^{-x}-1\right)}{2 \tan x-\arctan x}=$

$=\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 100}\right)^{x}-1\right)-\left(\left(e^{\ln 7}\right)^{-x}-1\right)}{2 \tan x-\arctan x}=$

$=\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 100}-1\right)-\left(e^{-x \ln 7}-1\right)}{2 \tan x-\arctan x}=$

$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 100}-1\right)-\left(e^{-x \ln 7}-1\right)\right)}{\frac{1}{x}(2 \tan x-\arctan x)}=$

$=\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 100}-1\right)-\left(e^{-x \ln 7}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(2 \tan x-\arctan x)}=$

$=\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 100}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-x \ln 7}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{2 \tan x}{x}-\lim _{x \rightarrow 0} \frac{\arctan x}{x}\right)=$

Using the substitution of equivalent infinitesimals:

$e^{x \ln 100}-1 \sim x \ln 100$, as $x \rightarrow 0(x \ln 100 \rightarrow 0)$

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

$\tan x \sim x$, as $x \rightarrow 0$

$\arctan x \sim x$, as $x \rightarrow 0$

We get:

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{x \ln 100}{x}-\lim _{x \rightarrow 0} \frac{-x \ln 7}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} \ln 100-\lim _{x \rightarrow 0}-\ln 7}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}= \\
& =\frac{\ln 100+\ln 7}{2-1}=\ln 700
\end{aligned}
$$

## Problem Kuznetsov Limits 15-15