## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0} \frac{\sin b x - \sin a x}{\ln \left(\tan\left(\frac{\pi}{4} + a x\right)\right)}
$$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin b x-\sin a x}{\ln \left(\tan\left(\frac{\pi}{4}+a x\right)\right)}=\lim _{x \rightarrow 0} \frac{2 \sin \frac{b x-a x}{2} \cos \frac{b x+a x}{2}}{\ln \left(\frac{\tan \frac{\pi}{4}+\tan a x}{1-\tan \frac{\pi}{4} \cdot \tan a x}\right)}= \\
& =\lim _{x \rightarrow 0} \frac{2 \sin \frac{b x-a x}{2} \cos \frac{b x+a x}{2}}{\ln \left(\frac{1+\tan a x}{1-\tan a x}\right)}=\lim _{x \rightarrow 0} \frac{2 \sin \frac{b x-a x}{2} \cos \frac{b x+a x}{2}}{\ln \left(\frac{1-\tan a x+2 \tan a x}{1-\tan a x}\right)}= \\
& =\lim _{x \rightarrow 0} \frac{2 \sin \frac{x(b-a)}{2} \cos \frac{x(b+a)}{2}}{\ln \left(1+\frac{2 \tan a x}{1-\tan a x}\right)}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& \sin \frac{x(b-a)}{2} \sim \frac{x(b-a)}{2} \quad, \quad x \rightarrow 0\left(\frac{x(b-a)}{2} \rightarrow 0\right) \\
& \ln \left(1+\frac{2 \tan a x}{1-\tan a x}\right) \sim \frac{2 \tan a x}{1-\tan a x}, \quad x \rightarrow 0\left(\frac{2 \tan a x}{1-\tan a x} \rightarrow 0\right)
\end{aligned}
$$

We get:

$$
=\lim _{x \rightarrow 0} \frac{2 \frac{x(b-a)}{2} \cos \frac{x(b+a)}{2}}{\frac{2 \tan a x}{1-\tan a x}}=\lim _{x \rightarrow 0} \frac{x(b-a)(1-\tan a x) \cos \frac{x(b+a)}{2}}{2 \tan a x}=
$$

Using the substitution of equivalent infinitesimals:

$\tan a x \sim a x$, as $x \rightarrow 0(a x \rightarrow 0)$

We get:

$$
=\lim _{x \rightarrow 0} \frac{x(b-a)(1-\tan a x) \cos \frac{x(b+a)}{2}}{2 a x}=
$$

$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{(b-a)(1-\tan a x) \cos \frac{x(b+a)}{2}}{2 a}= \\
& =\frac{(b-a)(1-\tan(a \cdot 0)) \cos \frac{0 \cdot(b+a)}{2}}{2 a}= \\
& =\frac{(b-a)(1-0) \cos 0}{2 a}=\frac{b-a}{2 a}
\end{aligned}
$$

## Problem Kuznetsov Limits 16-28