## Problem Statement

Calculate the limit of the function:
$\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin (x-h)}{h}$
## Solution

$\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin (x-h)}{h}=$

$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{(x+h)-(x-h)}{2} \cos \frac{(x+h)+(x-h)}{2}}{h}=$

$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{2 h}{2} \cos \frac{2 x}{2}}{h}=$

$=\lim _{h \rightarrow 0} \frac{2 \sin h \cos x}{h}=$

Using the substitution of equivalent infinitesimals:

$\sin h \sim h$, as $h \rightarrow 0$

We get:

$=\lim _{h \rightarrow 0} \frac{2 h \cos x}{h}=\lim _{h \rightarrow 0} 2 \cos x=2 \cos x$

## Problem Kuznetsov Limits 16-19