Renamed mathematical analysis to calculus

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-% Document MUST be built with XeLaTeX!
-\documentclass[11pt,a4paper]{article}
-
-\usepackage[margin=2.5cm]{geometry}
-
-% Поддръжка на български език
-% Thanks to https://tex.stackexchange.com/a/136926
-\usepackage{fontspec}
-\defaultfontfeatures{Ligatures=TeX}
-\setmainfont{CMU Serif}
-\setsansfont{CMU Sans Serif}
-
-\usepackage[bulgarian,english]{babel}
-
-\usepackage{amsmath}            % Математически символи
-\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. Вариант 2 за студентите с четна последна
-цифра от факултетния номер.}
-\author{Камен Димитров Младенов, Група 6}
-\date{10 Ноември 2021}
-
-\begin{document}
-
-\maketitle
-
-\paragraph{\colorbox{yellow}{Задача 1.}}
-
-Изследвайте функцията за обратимост. Намерете максимален интервал
-на обратимост, както и нейната обратна
-
-\begin{equation*}
-	f(x) = \log_{5} \left( x^2 - 2x + 1 \right)
-\end{equation*}
-
-\paragraph{Решение:}
-
-\begin{align*}
-	& f(x) = \log_{5} \left( x^2 - 2x + 1 \right) = \log_{5} \left( \left(x - 1\right) ^ 2 \right) \quad \text{Д.С. } x \neq 1 \text{ (аргумент на логаритъм)} \\
-	& \omit\hfill \\
-	& \text{Разглеждаме за инективност. Нека } f\left( x_1 \right) = f\left( x_2 \right) \qquad x_1, x_2 \in \mathbb{R} \setminus \{-1\} \\
-	& \log_{5} \left( \left(x_1 - 1\right) ^ 2 \right) = \log_{5} \left( \left(x_2 - 1\right) ^ 2 \right) \iff \left(x_1 - 1\right) ^ 2 = \left(x_2 - 1\right) ^ 2 \iff |x_1 - 1| = |x_2 - 1| \\
-	& \textbf{I. Случай: } x_1 - 1 > 0 \text{ и } x_2 - 1 > 0 \;,\; x_1 > 1 \quad x_2 > 1 \;,\; x_1,x_2 \in \left( 1, +\infty \right) \\
-	& \implies |x_1 - 1| = x_1 - 1 \text{ и } |x_2 - 1| = x_2 - 1 \\
-	& \text{Тоест } |x_1 - 1| = |x_2 - 1| \iff x_1 - 1 = x_2 - 1 \implies x_1 = x_2 \; \colorbox{yellow}{има инекция} \\
-	& \textbf{II. Случай: } x_1 - 1 < 0 \text{ и } x_2 - 1 < 0 \;,\; x_1 < 1 \quad x_2 < 1 \;,\; x_1,x_2 \in \left( -\infty, 1 \right) \\
-	& \implies |x_1 - 1| = 1 - x_1\text{ и } |x_2 - 1| = 1 - x_2 \\
-	& \text{Тоест } |x_1 - 1| = |x_2 - 1| \iff 1 - x_1 = 1 - x_2 \implies x_1 = x_2 \; \colorbox{yellow}{има инекция} \\
-	& \omit\hfill \\
-	& \text{Нека } y \in \mathbb{R} \text{. Тогава търсим x, такова че: } f(x) = y \\
-	& y = f(x) \quad y = \log_{5} \left( \left(x - 1\right) ^ 2 \right) \quad 5^y = \left(x - 1\right) ^ 2 \quad \sqrt{5^y} = |x - 1| \\
-	& \textbf{I. Случай: } x \in \left( 1, +\infty \right) \implies \sqrt{5^y} = x - 1 \quad x = 1 + \sqrt{5^y} \\
-	& x > 1 \quad 1 + \sqrt{5^y} > 1 \quad \sqrt{5^y} > 0 \quad \text{изпълнено за всяко y} \\
-	& \textbf{II. Случай: } x \in \left( -\infty, 1 \right) \implies \sqrt{5^y} = 1 - x \quad x = 1 - \sqrt{5^y} \\
-	& x < 1 \quad 1 - \sqrt{5^y} < 1 \quad - \sqrt{5^y} < 0 \quad \sqrt{5^y} > 0 \quad \text{изпълнено за всяко y} \\
-	& \omit\hfill \\
-	& \colorbox{yellow}{$f^{-1}: \mathbb{R} \rightarrow \left( -\infty, 1 \right) \cup \left( 1, +\infty \right) \qquad f^{-1}\left( y \right) = 1 \pm \sqrt{5^y}$}
-\end{align*}
-
-\hfill
-
-\paragraph{\colorbox{yellow}{Задача 2.}}
-
-Докажете тъждеството
-
-\begin{equation*}
-	\arcsin \left( 2x \sqrt{1-x^2} \right) = \pi - 2 \arccos \left( x \right) \text{, където } x \in \left[ \frac{-\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right]
-\end{equation*}
-
-\paragraph{Решение:}
-
-\begin{align*}
-	& \arcsin \left( 2x \sqrt{1-x^2} \right) = \alpha \iff \sin \left( \alpha \right) = 2x \sqrt{1-x^2} \qquad \arccos \left( x \right) = \beta \iff \cos \left( \beta \right) = x \\
-	& \text{За } x = \frac{-\sqrt{2}}{2} \;,\quad \sin(\alpha) = \cancel{2} \cdot \frac{-\sqrt{2}}{\cancel{2}} \cdot \sqrt{1 - \left( \frac{-\sqrt{2}}{2} \right)^2} = -\sqrt{2} \cdot \sqrt{\frac{1}{2}} = -1 \;,\; \cos \left( \beta \right) = -\frac{\sqrt{2}}{2} \\
-	& \text{Аналогично за } x = \frac{\sqrt{2}}{2} \;,\quad \sin \left( \alpha \right) = 1 \;,\quad \cos \left( \beta \right) = \frac{\sqrt{2}}{2} \\
-	& \implies \alpha \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \;,\; \beta \in \left[ -\frac{\pi}{4}, \frac{\pi}{4} \right] \implies \sin \left( \beta \right) = +\sqrt{1 - \cos ^ 2 \left( \beta \right)} = \sqrt{1 - x ^ 2} \\
-	& \omit\hfill \\
-	& \sin \left( \alpha \right) = 2x\sqrt{1 - x^2} = 2\cos \left( \beta \right)\sin \left( \beta \right) = \sin \left( 2\beta \right) = \sin \left( \pi - 2\beta \right) \qquad\quad 2\beta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \\
-	& \implies \colorbox{yellow}{$\alpha = \pi - 2\beta$}
-\end{align*}
-
-\hfill
-
-\paragraph{\colorbox{yellow}{Задача 3.}}
-
-Пресметнете границата на редицата
-
-\begin{equation*}
-	\lim_{n \to \infty} \frac{\sqrt[n-5]{n-5} + 2^{2n} + n^{12}}{10^n-n^3+(-1)^{3n + 5}}
-\end{equation*}
-
-\paragraph{Решение:}
-
-\begin{align*}
-	& \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 \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*}
-
-\hfill
-
-\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}
-\end{equation*}
-
-\paragraph{Решение:}
-
-\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}
diff --git a/Year 1/Mathematical analysis/Homeworks/Homework2.pdf b/Year 1/Mathematical analysis/Homeworks/Homework2.pdf
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diff --git a/Year 1/Mathematical analysis/Homeworks/Homework2.tex b/Year 1/Mathematical analysis/Homeworks/Homework2.tex
deleted file mode 100644
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-% Document MUST be built with XeLaTeX!
-\documentclass[11pt,a4paper]{article}
-
-\usepackage[margin=1.5cm]{geometry}
-
-% Поддръжка на български език
-% Thanks to https://tex.stackexchange.com/a/136926
-\usepackage{fontspec}
-\defaultfontfeatures{Ligatures=TeX}
-\setmainfont{CMU Serif}
-\setsansfont{CMU Sans Serif}
-
-\usepackage[bulgarian,english]{babel}
-
-\usepackage{amsmath}            % Математически символи
-\usepackage{extarrows}
-\usepackage{amsfonts}           % Още математически символи
-\usepackage{cancel}             % Зачертаване
-\usepackage[dvipsnames]{xcolor} % Цветни кутийки + допълнителни цветове
-\newcommand{\answerbox}[1]{\colorbox{yellow}{$\displaystyle #1$}}
-
-\usepackage{tikz}
-\newcommand{\lipoint}[3]{\draw (#1, 3pt) -- (#1, -3pt) node[below] {#2} node[above] {#3}}
-\newcommand{\parch}[2]{\node[above] at (#1, 0) {#2} ; \draw (#1 + 0.5, 0) arc (0:180:0.5)}
-
-\usepackage{enumitem}
-\setlist[enumerate]{label=\arabic*.}
-
-\usepackage[pdfauthor={Камен Димитров Младенов, КН Зимен семестър, 2021},
-		    pdftitle={Домашна работа 2 по ДИС 1. Компютърни науки, 1 курс, 2 поток},
-		    pdfsubject={},
-		    pdfkeywords={},
-		    colorlinks=true,
-		    urlcolor=blue]{hyperref}
-
-\title{\vspace{-1.5cm}Домашна работа 2 по ДИС 1. Компютърни науки, 1 курс, 2 поток}
-\author{Камен Димитров Младенов, Група 6}
-\date{20 Ноември 2021}
-
-\begin{document}
-
-\maketitle
-
-\noindent
-\colorbox{YellowGreen}{\parbox{\textwidth}{\; Отличен 5,75/6 \qquad [Коментарите са под определените места]}}
-
-\paragraph{\colorbox{yellow}{Задача 1.}}
-
-Дадена е редицата \hfill \colorbox{Dandelion}{0,6/0,8 т.}
-\begin{align*}
-	a_1 = \lambda, \; a_{n+1} = \frac{2a_n^2 - a_n + 6}{a_n - 6}
-\end{align*}
-Да се изследва за сходимост спрямо стойностите на параметъра $\lambda \neq 6$.
-
-\paragraph{Решение:}
-
-Разглеждаме за монотонност:
-\begin{align*}
-	& a_{n+1} - a_n = \frac{2a_n^2 - a_n + 6}{a_n - 6} - a_n = \frac{2a_n^2 - a_n + 6 -a_n^2 + 6a_n}{a_n - 6} = \frac{a_n^2 + 5a_n + 6}{a_n - 6} = \frac{(a_n+3)(a_n+2)}{a_n - 6} \\
-	& \implies \text{Корените за } a_n \text{ са -3, -2 и 6}
-\end{align*}
-\hfill
-\begin{tikzpicture}
-	\draw[-latex] (-0.25, 0) -- (4.25, 0) ;
-	\parch{3.5}{+} ;
-	\lipoint{3}{6}{} ;
-	\parch{2.5}{-} ;
-	\lipoint{2}{-2}{} ;
-	\parch{1.5}{+};
-	\lipoint{1}{-3}{} ;
-	\parch{0.5}{-} ;
-\end{tikzpicture}
-
-Разглеждаме положението на $a_{n+1}$ спрямо -3, -2 и 6
-
-\noindent
-\begin{align*}
-	& a_{n+1} - (-3) = \frac{2a_n^2 - a_n + 6}{a_n - 6} + 3 = \frac{2a_n^2 - a_n + 6 + 3a_n - 18}{a_n - 6} = \frac{2a_n^2 + 2a_n - 12}{a_n - 6} \\
-	& \implies \text{Корените за } a_n \text{ са -3, 2 и 6}
-\end{align*}
-\hfill
-\begin{tikzpicture}
-	\draw[-latex] (-0.25, 0) -- (4.25, 0) ;
-	\parch{3.5}{+} ;
-	\lipoint{3}{6}{} ;
-	\parch{2.5}{-} ;
-	\lipoint{2}{2}{} ;
-	\parch{1.5}{+};
-	\lipoint{1}{-3}{} ;
-	\parch{0.5}{-} ;
-\end{tikzpicture}
-
-\noindent
-\begin{align*}
-	& a_{n+1} - (-2) = \frac{2a_n^2 - a_n + 6}{a_n - 6} + 2 = \frac{2a_n^2 - a_n + 6 + 2a_n - 12}{a_n - 6} = \frac{2a_n^2 + a_n - 6}{a_n - 6} \\
-	& \implies \text{Корените за } a_n \text{ са -2, $\frac{3}{2}$ и 6}
-\end{align*}
-\hfill
-\begin{tikzpicture}
-	\draw[-latex] (-0.25, 0) -- (4.25, 0) ;
-	\parch{3.5}{+} ;
-	\lipoint{3}{6}{} ;
-	\parch{2.5}{-} ;
-	\lipoint{2}{$\frac{3}{2}$}{} ;
-	\parch{1.5}{+};
-	\lipoint{1}{-2}{} ;
-	\parch{0.5}{-} ;
-\end{tikzpicture}
-
-\noindent
-\begin{align*}
-	& a_{n+1} - 6 = \frac{2a_n^2 - a_n + 6}{a_n - 6} - 6 = \frac{2a_n^2 - a_n + 6 - 6a_n + 36}{a_n - 6} = \frac{2a_n^2 - 7a_n + 42}{a_n - 6} \genfrac..{0pt}{0}{\; >0}{} \\
-	& \implies \text{Единствения корен за } a_n \text{ е 6}
-\end{align*}
-\hfill
-\begin{tikzpicture}
-	\draw[-latex] (-0.25, 0) -- (2.25, 0) ;
-	\parch{1.5}{+};
-	\lipoint{1}{6}{} ;
-	\parch{0.5}{-} ;
-\end{tikzpicture}
-
-Разглеждаме положението на $a_{n+1}$ спрямо $\frac{3}{2}$ и 2
-
-\noindent
-\begin{align*}
-	& a_{n+1} - \frac{3}{2} = \frac{2a_n^2 - a_n + 6}{a_n - 6} - \frac{3}{2} = \frac{2a_n^2 - a_n + 6 - \frac{3}{2}a_n + 9}{a_n - 6} = \frac{2a_n^2 - \frac{5}{2}a_n + 15}{a_n - 6} \genfrac..{0pt}{0}{\; >0}{} \\
-	& \implies \text{Единствения корен за } a_n \text{ е 6}
-\end{align*}
-
-\noindent
-\begin{align*}
-	& a_{n+1} - 2 = \frac{2a_n^2 - a_n + 6}{a_n - 6} - 2 = \frac{2a_n^2 - a_n + 6 - 2a_n + 12}{a_n - 6} = \frac{2a_n^2 - 3a_n + 18}{a_n - 6} \genfrac..{0pt}{0}{\; >0}{} \\
-	& \implies \text{Единствения корен за } a_n \text{ е 6}
-\end{align*}
-
-Разглеждаме положението на $a_{n+1}$
-
-\noindent
-\begin{align*}
-	& a_{n+1} = \frac{2a_n^2 - a_n + 6}{a_n - 6} \genfrac..{0pt}{0}{\; >0}{} \implies \text{Единствения корен за } a_n \text{ е 6}
-\end{align*}
-
-\begin{center}
-	\begin{tabular}{c|ccccccccccc}
-		$a_n$ & $-\infty$ & -3 & & -2 & & $\frac{3}{2}$ & & 2 & & 6 & $+\infty$ \\
-		\hline
-		$a_{n+1} - a_n$ & - & 0 & + & 0 & - & \vdots & - & \vdots & - & 0 & + \\
-		$a_{n+1} - (-3)$ & - & 0 & + & \vdots & + & \vdots & + & 0 & - & 0 & + \\
-		$a_{n+1} - (-2)$ & - & \vdots & - & 0 & + & 0 & - & \vdots & - & 0 & + \\
-		$a_{n+1}$ & - & \vdots & - & \vdots & - & \vdots & - & \vdots & - & 0 & + \\
-		$a_{n+1} - \frac{3}{2}$ & - & \vdots & - & \vdots & - & \vdots & - & \vdots & - & 0 & + \\
-		$a_{n+1} - 2$ & - & \vdots & - & \vdots & - & \vdots & - & \vdots & - & 0 & + \\
-		$a_{n+1} - 6$ & - & \vdots & - & \vdots & - & \vdots & - & \vdots & - & 0 & + \\
-	\end{tabular}
-\end{center}
-
-\begin{enumerate}[label={\textbf{\arabic*. Случай}},leftmargin=5.8em]
-	\item $\lambda < -3$ \quad В този интервал $a_{n+1} < a_n$, следователно $a_{n+1} < -3$ и редицата е монотонно намаляваща, но не е ограничена отдолу. Тоест $\{ a_n \}_{n=1}^{\infty}$ не е сходяща.
-	\colorbox{RedOrange}{дивергира към $-\infty$}
-	
-	\item $\lambda = -3$ \quad Пресмятаме, че $a_2 = -3$ и $a_3 = -3$. Ще докажем по индукция, че $a_n = -3$:
-	\begin{itemize}
-		\item \emph{База:} Вярно за $a_1, a_2, a_3$
-		\item \emph{Индукционно предположение:} $a_{n-1} = -3$ за някое $n \geq 1$.
-		\item \emph{Индукционна стъпка:} $a_n \xlongequal{def} \frac{2a_{n-1}^2 - a_{n-1} + 6}{a_{n-1} - 6} \xlongequal{\text{ИП}} \frac{2 \cdot 9 + 3 + 6}{-3 -6} = -3$
-	\end{itemize}
-	То тогава редицата е стационарна и следователно сходяща.
-
-	\item $\lambda \in (-3, -2)$ \quad Пресмятаме, че $a_2 \in (-3; -2)$ и $a_3 \in (-3; -2)$. Ще докажем по индукция, че $a_n \in (-3; -2)$:
-	\begin{itemize}
-		\item \emph{База:} Вярно за $a_1, a_2, a_3$
-		\item \emph{Индукционно предположение:} $a_{n-1} \in (-3; -2)$ за някое $n \geq 1$.
-		\item \emph{Индукционна стъпка:} $a_n \xlongequal{def} \frac{2a_{n-1}^2 - a_{n-1} + 6}{a_{n-1} - 6} \stackrel{\text{ИП}}{>} \frac{2(-3)^2 - (-3) + 6}{(-3) - 6} = -3$ \newline
-		$\- \qquad \frac{2a_{n-1}^2 - a_{n-1} + 6}{a_{n-1} - 6} \stackrel{\text{ИП}}{<} \frac{2(-2)^2 - (-2) + 6}{(-2) - 6} = -2$
-		\colorbox{RedOrange}{не е достатъчно само да е ограничена: монотонна е}
-	\end{itemize}
-	То тогава редицата е ограничена отдолу и отгоре. Следователно тя е сходяща. \colorbox{RedOrange}{към?}
-	
-	\item $\lambda = -2$ \quad Пресмятаме, че $a_2 = -2$ и $a_3 = -2$. По индукция, следва че редицата е стационарна и следователно сходяща.	
-	
-	\item $\lambda \in \left(-2, \frac{3}{2}\right)$ \quad Пресмятаме, че $a_1, a_2 > -2$. Знаем, че $a_{n+1} < \frac{3}{2}$, тоест по индукция намираме, че $a_n \in \left(-2; \frac{3}{2}\right)$. То редицата е огрничена и следователно сходяща. \colorbox{Dandelion}{аналогично на 3.}
-
-	\item $\lambda = \frac{3}{2}$ \quad Пресмятаме, че $a_2 = -2$ и $a_3 = -2$. По индукция, следва че редицата е стационарна и следователно сходяща.
-	
-	\item $\lambda \in \left(\frac{3}{2}, 2\right)$ \quad Пресмятаме, че $a_2 \in (-3; -2)$ и $a_3 \in (-3; -2)$ и по индукция $a_n \in (-3; -2)$. Тоест редицата е ограничена и следователно сходяща. \colorbox{Dandelion}{аналогично на 3.}
-	
-	\item $\lambda = 2$ \quad Пресмятаме, че $a_2 = -3$ и $a_3 = -3$. По индукция, следва че редицата е стационарна и следователно сходяща.
-	
-	\item $\lambda \in (2, 6)$ \quad По индукция намираме, че $a_n < -3$. $a_{n+1} < a_n$, следователно редицата е монотонно намаляваща, но не е ограничена отдолу. Тоест не е сходяща. \colorbox{RedOrange}{дивергира към $-\infty$}
-	
-	\item $\lambda > 6$ \quad По индукция намираме, че $a_n > 6$. Знаем, че $a_{n+1} > a_n$, следователно редицата е монотонно растяща, но не е ограничена отгоре. Тоест не е сходяща. \colorbox{RedOrange}{дивергира към $+\infty$}
-\end{enumerate}
-
-\paragraph{\colorbox{yellow}{Задача 2.}}
-
-Да се пресметне границата \hfill \colorbox{YellowGreen}{0,8-/0,8 т.}
-\begin{align*}
-	\lim_{x \rightarrow \infty} \frac{\sqrt{2x^2 + 1} - \sqrt[3]{4x^2 - 1}}{\sqrt[4]{x^4 + 7} - \sqrt[5]{x^4 - 3}}
-\end{align*}
-
-\paragraph{Решение:}
-\begin{align*}
-	\lim_{x \rightarrow \infty} \frac{\sqrt{2x^2 + 1} - \sqrt[3]{4x^2 - 1}}{\sqrt[4]{x^4 + 7} - \sqrt[5]{x^4 - 3}} = \lim_{x \rightarrow \infty} \frac{\cancel{x} \sqrt{2 + \frac{1}{x^2}} - \cancel{x} \sqrt[3]{\frac{4}{x} - \frac{1}{x^3}}}{\cancel{x} \sqrt[4]{1 + \frac{7}{x^4}} - \cancel{x} \sqrt[5]{\frac{1}{x} - \frac{3}{x^4}}} = L
-	\quad \colorbox{RedOrange}{под последния корен е $x^5$ вместо $x^4$}
-\end{align*}
-\begin{align*}
-	& \sqrt{2} \xrightarrow[x \rightarrow \infty]{} \sqrt{2} \;,\; \frac{1}{x^2} \xrightarrow[x \rightarrow \infty]{} 0 \;,\; \frac{4}{x} \xrightarrow[x \rightarrow \infty]{} 0 \;,\; \frac{1}{x^3} \xrightarrow[x \rightarrow \infty]{} 0 \;,\; \\
-	& \sqrt[4]{1} \xrightarrow[x \rightarrow \infty]{} \sqrt[4]{1} \;,\; \frac{7}{x^4} \xrightarrow[x \rightarrow \infty]{} 0 \;,\; \frac{1}{x} \xrightarrow[x \rightarrow \infty]{} 0 \;,\; \frac{3}{x^4} \xrightarrow[x \rightarrow \infty]{} 0
-	\qquad 	\implies L = \frac{\sqrt{2}}{\sqrt[4]{1}} = \answerbox{\sqrt{2}}
-\end{align*}
-
-\paragraph{\colorbox{yellow}{Задача 3.}}
-
-Да се пресметне границата \hfill \colorbox{YellowGreen}{0,8/0,8 т.}
-\begin{align*}
-	\lim_{x \rightarrow 0} \frac{\sqrt[3]{5x + 1} - \sqrt[4]{7x + 1}}{\arctg(\ln(x+1))}
-\end{align*}
-
-\paragraph{Решение:}
-
-\begin{align*}
-	\arctg(\ln(x+1)) \sim \ln(x+1) \quad x \rightarrow 0 \qquad \sqrt[3]{5x + 1} = \underbrace{(1 + 5x)^{\frac{1}{3}} - 1}_{\sim \frac{5x}{3} \quad x \rightarrow 0} + 1 \qquad \sqrt[4]{7x + 1} = \underbrace{(1 + 7x)^{\frac{1}{4}} - 1}_{\sim \frac{7x}{4} \quad x \rightarrow 0} + 1
-\end{align*}
-\begin{align*}
-	\implies \lim_{x \rightarrow 0} \frac{\sqrt[3]{5x + 1} - \sqrt[4]{7x + 1}}{\arctg(\ln(x+1))} \xlongequal{(\sim)} \lim_{x \rightarrow 0} \frac{\frac{5x}{3} + 1 - \left( \frac{7x}{4} + 1 \right)}{\ln(x+1)} = \lim_{x \rightarrow 0} \frac{\frac{20x - 21x}{12}}{\ln(x+1)} = \lim_{x \rightarrow 0} - \frac{x}{12 \ln(x+1)} = L
-\end{align*}
-\begin{align*}
-	\frac{x}{\ln(x+1)} \xrightarrow[x \rightarrow 0]{} 1 \implies L = \answerbox{- \frac{1}{12}}
-\end{align*}
-
-\paragraph{\colorbox{yellow}{Задача 4.}}
-
-Да се пресметне $f^{(n)}(x)$, където \hfill \colorbox{YellowGreen}{0,8/0,8 т.}
-\begin{align*}
-	f(x) = \frac{x^3}{\sqrt{1 - 7x}}
-\end{align*}
-
-\paragraph{Решение:}
-
-\begin{align*}
-	g(x) = \frac{1}{\sqrt{1 - 7x}} = (1 - 7x)^{-\frac{1}{2}} \qquad h(x) = x^3 \qquad\qquad \implies f(x) = g(x) \cdot h(x)
-\end{align*}
-\begin{align*}
-	g^{(n)}(x) = \left( (1 - 7x)^{-\frac{1}{2}} \right)^{(n)} & = \left(-\frac{1}{2} \right) \left( -\frac{1}{2} - 1 \right) \left( -\frac{1}{2} - 2 \right) \cdots \left( -\frac{1}{2} - n + 1 \right) \cdot x^{-\frac{1}{2} - n} \cdot 7^n \cdot (1-7x)^{-\frac{2n+1}{2}} = \\
-	& = \frac{(-1)(-3)(-5) \cdots (-1-2n+2)}{2^n} \cdot x^{-\frac{1}{2} - n} \cdot 7^n \cdot (1-7x)^{-\frac{2n+1}{2}} = \\
-	& = \frac{(-1)(-3) \cdots (1-2n)}{2^n} \cdot x^{-\frac{1+2n}{2}} \cdot 7^n \cdot (1-7x)^{-\frac{2n+1}{2}}
-\end{align*}
-\begin{align*}
-	h^{(0)}(x) = h(x) = x^3 \quad h'(x) = 3x^2 \quad h''(x) = 6x \quad h'''(x) = 6 \quad h^{IV}(x) = 0 \quad \implies h^{(n)}(x) = 0 \quad n \geq 4
-\end{align*}
-\begin{align*}
-	f^{(n)} & = \sum_{k = 0}^n g^{(n-k)} h^{(k)} = \binom{n}{0}g^{(n)}h^{(0)} + \binom{n}{1}g^{(n-1)}h^{(1)} + \binom{n}{2}g^{(n-2)}h^{(2)} + \binom{n}{3}g^{(n-3)}h^{(3)} + \\
-	& \quad + \underbrace{\binom{n}{4}g^{(n-4)}h^{(4)} + \cdots + \binom{n}{0}g^{(0)}h^{(n)}}_{0} = \\
-	& = 1 \cdot \frac{(-1)(-3) \cdots (1-2n)}{2^n} \; x^{-\frac{1+2n}{2}} \; 7^n \; (1-7x)^{-\frac{2n+1}{2}} \cdot x^3 \\
-	& \quad + n \cdot \frac{(-1)(-3) \cdots (3-2n)}{2^{n-3}} \; x^{-\frac{2n-1}{2}} \; 7^{n-1} \; (1-7x)^{-\frac{2n-1}{2}} \cdot 3x^2 \\
-	& \quad + \frac{n!}{2(n-2)!} \cdot \frac{(-1)(-3) \cdots (5-2n)}{2^{n-2}} \; x^{-\frac{2n - 3}{2}} \; 7^{n-5} \; (1-7x)^{-\frac{2n-2}{2}} \cdot 6x \\
-	& \quad + \frac{n!}{6(n-3)!} \cdot \frac{(-1)(-3) \cdots (7-2n)}{2^{n-3}} \; x^{-\frac{2n-5}{2}} \; 7^{n-3} \; (1-7x)^{-\frac{2n-7}{2}} \cdot 6 = \\
-	& = \frac{(-1)(-3) \cdots (1-2n)}{2^n} \; x^{-\frac{2n-5}{2}} \; 7^n \; (1-7x)^{-\frac{2n+1}{2}} \\
-	& \quad + 3n \cdot \frac{(-1)(-3) \cdots (3-2n)}{2^{n-1}} \; x^{-\frac{2n-5}{2}} \; 7^{n-1} \; (1-7x)^{-\frac{2n-1}{2}} \\
-	& \quad + 3\frac{n!}{(n-2)!} \cdot \frac{(-1)(-3) \cdots (5-2n)}{2^{n-2}} \; x^{-\frac{2n-7}{2}} \; 7^{n-5} \; (1-7x)^{-\frac{2n-2}{2}} \\
-	& \quad + \frac{n!}{(n-3)!} \cdot \frac{(-1)(-3) \cdots (7-2n)}{2^{n-3}} \; x^{-\frac{2n-2}{2}} \; 7^{n-3} \; (1-7x)^{-\frac{2n-7}{2}}
-\end{align*}
-
-\paragraph{\colorbox{yellow}{Задача 5.}}
-
-Да се пресметне границата \hfill \colorbox{YellowGreen}{0,8-/0,8 т.}
-\begin{align*}
-	\lim_{x \rightarrow 0} \left( \frac{\sin x}{\arcsin x} \right) ^ {\frac{1}{x^2}}
-\end{align*}
-
-\paragraph{Решение:}
-
-\begin{align*}
-	\lim_{x \rightarrow 0} \left( \frac{\sin x}{\arcsin x} \right) ^ {\frac{1}{x^2}} = \lim_{x \rightarrow 0} e^{\ln\left( \frac{\sin x}{\arcsin x} \right) ^ {\frac{1}{x^2}}} = \lim_{x \rightarrow 0} e^{\frac{1}{x^2} \cdot \ln\left( \frac{\sin x}{\arcsin x} \right)} = e^l
-\end{align*}
-\begin{align*}
-	l & = \lim_{x \rightarrow 0} \frac{1}{x^2} \ln\left( \frac{\sin x}{\arcsin x} \right) \qquad \ln\left( \frac{\sin x}{\arcsin x} - 1 + 1 \right) \sim \frac{\sin x}{\arcsin x} - 1 \quad x \rightarrow 0 \quad \colorbox{RedOrange}{$\frac{\sin x}{\arcsin x} \rightarrow 1$ при $x \rightarrow 1$} \\
-	& \implies l \xlongequal{(\sim)} \lim_{x \rightarrow 0} \frac{1}{x^2} \left(  \frac{\sin x}{\arcsin x} - 1 \right) = \lim_{x \rightarrow 0} \frac{1}{x^2} \left(  \frac{\sin x - \arcsin x}{\arcsin x} \right) \qquad \arcsin x \sim x \quad x \rightarrow 0 \\
-	& \implies l \xlongequal{(\sim)} \lim_{x \rightarrow 0} \frac{1}{x^2} \left(  \frac{\sin x - \arcsin x}{x} \right) = \lim_{x \rightarrow 0} \frac{\sin x - \arcsin x}{x^3} \xlongequal{\text{Лопитал } \left[ \frac{0}{0} \right]} \lim_{x \rightarrow 0} \frac{\cos x - \frac{1}{\sqrt{1-x^2}}}{3x^2} \\
-	& \qquad \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{-\frac{1}{2}} \quad \left( (1-x^2)^{-\frac{1}{2}} \right)' = -\frac{1}{2} (1-x^2)^{-\frac{3}{2}} \cdot \left( (1-x^2) \right)' = x(1-x^2)^{-\frac{3}{2}} \\
-	& \implies l \xlongequal{\text{Лопитал } \left[ \frac{0}{0} \right]} \lim_{x \rightarrow 0} \frac{-\sin x - x(1-x^2)^{-\frac{3}{2}}}{6x} \\
-	& \qquad \left( x(1-x^2)^{-\frac{3}{2}} \right)' = \left( \frac{x}{(1-x^2)^{\frac{3}{2}}} \right)' = \frac{(1-x^2)^{\frac{3}{2}} + \frac{3}{2}(1-x^2)^{\frac{1}{2}} 2x \cdot x}{(1-x^2)^3} = \frac{(1-x^2)\sqrt{1-x^2} + 3x^2\sqrt{1-x^2}}{(1-x^2)^3} = \\
-	& \qquad = \frac{2x^2+1}{(1-x^2)^{\frac{5}{2}}} \\
-	& \implies l \xlongequal{\text{Лопитал } \left[ \frac{0}{0} \right]} \lim_{x \rightarrow 0} \frac{-\cos x - \frac{2x^2+1}{(1-x^2)^{\frac{5}{2}}}}{6} \qquad \cos x \xrightarrow[x \rightarrow 0]{} 1 \qquad \frac{2x^2+1}{(1-x^2)^{\frac{5}{2}}} \xrightarrow[x \rightarrow 0]{} 1 \quad \implies l = -\frac{1}{3}
-\end{align*}
-
-Следователно, търсената граница е: $e^l = e^{-\frac{1}{3}} = \answerbox{\frac{1}{\sqrt[3]{e}}}$
-
-\end{document}
\ No newline at end of file