Improved Ex. 3 solution and updated date

2 files changed, 3 insertions, 3 deletions

diff --git a/Year 1/Mathematical analysis/Homework1.pdf b/Year 1/Mathematical analysis/Homework1.pdf
index d55e677..4c4694a 100644
--- a/Year 1/Mathematical analysis/Homework1.pdf
+++ b/Year 1/Mathematical analysis/Homework1.pdf
diff --git a/Year 1/Mathematical analysis/Homework1.tex b/Year 1/Mathematical analysis/Homework1.tex
index 08027dc..08a137a 100644
--- a/Year 1/Mathematical analysis/Homework1.tex
+++ b/Year 1/Mathematical analysis/Homework1.tex
@@ -13,14 +13,14 @@
 \usepackage[bulgarian,english]{babel}
 
 \usepackage{amsmath}            % Математически символи
-\usepackage{amsfonts}
+\usepackage{amsfonts}           % Още математически символи
 \usepackage{hyperref}           % Линкове
 \usepackage{cancel}             % Зачертаване
 \usepackage[dvipsnames]{xcolor} % Цветни кутийки + допълнителни цветове
 
 \title{\vspace{-2cm}Домашна работа 1 по ДИС 1, за спец. КН 1 курс, 2 поток}
 \author{Камен Димитров Младенов, Група 6}
-\date{14 Ноември 2021}
+\date{10 Ноември 2021}
 
 \begin{document}
 
@@ -97,7 +97,7 @@
 	& \lim_{n \to \infty} \frac{\sqrt[n-5]{n-5} + 2^{2n} + n^{12}}{10^n-n^3+(-1)^{3n + 5}} = \lim_{n \to \infty} \frac{2^{2n} \cdot \left( \frac{\sqrt[n-5]{n-5}}{2^{2n}} + 1 + \frac{n^{12}}{2^{2n}} \right)}{10^n \cdot \left(1 - \frac{n^3}{10^n} + \frac{(-1)^{3n + 5}}{10^n} \right)} = a_n \\
 	& \omit\hfill \\
 	& \lim_{n \to \infty} \frac{\sqrt[n-5]{n-5}}{2^{2n}} = \frac{\lim\limits_{n \to \infty} \sqrt[n-5]{n-5}}{ \lim\limits_{n \to \infty} 4^n} = \frac{1}{\infty} = 0 \;,\quad \lim_{n \to \infty} \frac{n^{12}}{2^{2n}} = 0 \;,\quad \lim_{n \to \infty} \frac{n^3}{10^n} = 0 \;,\; \\
-	& \lim_{n \to \infty} \frac{(-1)^{3n + 5}}{10^n} = \frac{\lim\limits_{n \to \infty} (-1)^{3n + 5}}{\lim\limits_{n \to \infty} 10^n} = \frac{\lim\limits_{n \to \infty} (-1)^{3n + 5}}{\infty} = 0 \; \left( \lim\limits_{n \to \infty} (-1)^{3n + 5} < \infty \right) \\
+	& \lim_{n \to \infty} \frac{(-1)^{3n + 5}}{10^n} = \frac{\lim\limits_{n \to \infty} (-1)^{3n + 5}}{\lim\limits_{n \to \infty} 10^n} = \frac{\lim\limits_{n \to \infty} (-1)^{3n + 5}}{\infty} = 0 \quad \text{($(-1)^{3n + 5}$ е разходяща)} \\
 	& \omit\hfill \\
 	& \implies a_n = \lim_{n \to \infty} \frac{2^{2n}}{10^n} = \lim_{n \to \infty} \frac{4^n}{10^n} = \lim_{n \to \infty} \left( \frac{2}{5} \right) ^ n \qquad \frac{2}{5} < 1 \implies \lim_{n \to \infty} \left( \frac{2}{5} \right) ^ n = \colorbox{yellow}{0}
 \end{align*}