## Problem Statement

Calculate the limit of the function:
$\lim _{h \rightarrow 0} \frac{\ln (x+h)+\ln (x-h)-2 \ln x}{h^{2}}, x>0$
## Solution

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

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1-\frac{h^{2}}{x^{2}}\right) \sim-\frac{h^{2}}{x^{2}}$, as $\quad h \rightarrow 0\left(-\frac{h^{2}}{x^{2}} \rightarrow 0\right)$

We get:

$=\lim _{h \rightarrow 0} \frac{-\frac{h^{2}}{x^{2}}}{h^{2}}=\lim _{h \rightarrow 0}-\frac{1}{x^{2}}=-\frac{1}{x^{2}}$

## Problem Kuznetsov Limits 16-15