Solved 4th exercise and updated some styling and package usage

2 files changed, 26 insertions, 10 deletions

diff --git a/Year 1/Mathematical analysis/Homework1.pdf b/Year 1/Mathematical analysis/Homework1.pdf
index c35bc40..0dbeea5 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 9572f46..a313856 100644
--- a/Year 1/Mathematical analysis/Homework1.tex
+++ b/Year 1/Mathematical analysis/Homework1.tex
@@ -1,7 +1,7 @@
 % Document MUST be built with XeLaTeX!
 \documentclass[11pt,a4paper]{article}
 
-\usepackage[left=3.5cm, right=3.5cm]{geometry}
+\usepackage[left=2.5cm, right=2.5cm]{geometry}
 
 % Поддръжка на български език
 % Thanks to https://tex.stackexchange.com/a/136926
@@ -12,8 +12,10 @@
 
 \usepackage[bulgarian,english]{babel}
 
-\usepackage{amsmath} % Математически символи
-\usepackage{hyperref} % Линкове
+\usepackage{amsmath}            % Математически символи
+\usepackage{hyperref}           % Линкове
+\usepackage{cancel}             % Зачертаване
+\usepackage[dvipsnames]{xcolor} % Цветни кутийки + допълнителни цветове
 
 \title{\vspace{-3cm}Домашна работа 1 по ДИС 1, за спец. КН 1 курс, 2 поток}
 \author{Камен Димитров Младенов, Група 6}
@@ -23,7 +25,7 @@
 
 \maketitle
 
-\paragraph{Задача 1.}
+\paragraph{\colorbox{yellow}{Задача 1.}}
 
 Изследвайте функцията за обратимост. Намерете максимален интервал
 на обратимост, както и нейната обратна
@@ -38,7 +40,7 @@
 
 \hfill
 
-\paragraph{Задача 2.}
+\paragraph{\colorbox{yellow}{Задача 2.}}
 
 Докажете тъждеството
 
@@ -52,7 +54,7 @@
 
 \hfill
 
-\paragraph{Задача 3.}
+\paragraph{\colorbox{yellow}{Задача 3.}}
 
 Пресметнете границата на редицата
 
@@ -66,16 +68,30 @@
 
 \hfill
 
-\paragraph{Задача 4.}
+\paragraph{\colorbox{yellow}{Задача 4.}}
 
 Пресметнете границата на редицата.
 
 \begin{equation*}
-	\lim_{n \to \infty} \left( \frac{n^2 - 5n + 14}{n^2 + 1} \right)^{\frac{n^2-n+2}{n+5}}
+	\lim_{n \to \infty} \left( \frac{n^2 - 5n + 14}{n^2 + 1} \right) ^ \frac{n^2-n+2}{n+5}
 \end{equation*}
 
 \paragraph{Решение}
 
-[TODO]
+\begin{align*}
+	& \lim_{n \to \infty} \left( \frac{n^2 - 5n + 14}{n^2 + 1} \right) ^ \frac{n^2-n+2}{n+5} = \lim_{n \to \infty} \left( \frac{n^2 + 1 - 5n + 13}{n^2 + 1} \right) ^ \frac{n^2-n+2}{n+5} = \\
+	& = \lim_{n \to \infty} \left( 1 + \frac{13 - 5n}{n^2 + 1} \right) ^ \frac{n^2-n+2}{n+5} = \lim_{n \to \infty} \left( 1 + \frac{1}{\frac{n^2 + 1}{13 - 5n}} \right) ^ { \frac{n^2+1}{13-5n} \cdot \frac{13-5n}{n^2+1} \cdot \frac{n^2-n+2}{n+5} }
+\end{align*}
+
+От $\lim\limits_{n \to \infty} \left( 1 + \frac{1}{p_n} \right) ^ {p_n b_n} = e ^ {\lim_{n \to \infty} b_n}$, задачата се свежда до $e ^ {\lim_{n \to \infty} b_n} = \; ? \;$, където
+
+$ \; b_n = \frac{13-5n}{n^2+1} \cdot \frac{n^2-n+2}{n+5}$
+
+\begin{align*}
+	\lim_{n \to \infty} b_n &= \lim_{n \to \infty} \frac{13-5n}{n^2+1} \cdot \frac{n^2-n+2}{n+5} = \lim_{n \to \infty} \frac{13n^2 - 13n + 26 - 5n^3 + 5n^2 - 10n}{n^3 + 5n^2 + n + 5} = \\
+	& = \lim_{n \to \infty} \frac{-5n^3 + 18n^2 - 23n + 26}{n^3 + 5n^2 + n + 5} = \lim_{n \to \infty} \frac{\cancel{n^3} \cdot \left( -5 + \frac{18}{n} - \frac{23}{n^2} + \frac{26}{n^3} \right)}{\cancel{n^3} \cdot \left( 1 + \frac{5}{n} + \frac{1}{n^2} + \frac{5}{n^3} \right)} = \\
+	& = \frac{\lim\limits_{n \to \infty} -5 + \lim\limits_{n \to \infty} \frac{18}{n} - \lim\limits_{n \to \infty} \frac{23}{n^2} + \lim\limits_{n \to \infty} \frac{26}{n^3}}{\lim\limits_{n \to \infty} 1 + \lim\limits_{n \to \infty} \frac{5}{n} + \lim\limits_{n \to \infty} \frac{1}{n^2} + \lim\limits_{n \to \infty} \frac{5}{n^3}} = \frac{-5+0+0+0}{1+0+0+0} = -5 \\
+	& \implies e ^ {\lim_{n \to \infty} b_n} = e ^ {-5} = \colorbox{yellow}{$\frac{1}{e ^ 5}$}
+\end{align*}
 
-\end{document}
\ No newline at end of file
+\end{document}