## Problem Statement

Calculate the limit of the function:

$\lim _{h \rightarrow 0} \frac{a^{x+h}+a^{x-h}-2 a^{x}}{h}$
Solution

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

$=\lim _{h \rightarrow 0} \frac{a^{x+h}-a^{x}}{h}+\lim _{h \rightarrow 0} \frac{a^{x-h}-a^{x}}{h}=$

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

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

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

Using the substitution of equivalent infinitesimals:

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

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

We get:

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

## Problem Kuznetsov Limits 16-21