Solved 5th exercise and almost completed ex1 of dis hw2

2 files changed, 34 insertions, 10 deletions

diff --git a/Year 1/Mathematical analysis/Homeworks/Homework2.pdf b/Year 1/Mathematical analysis/Homeworks/Homework2.pdf
index 4fe15f5..9295ac4 100644
--- a/Year 1/Mathematical analysis/Homeworks/Homework2.pdf
+++ b/Year 1/Mathematical analysis/Homeworks/Homework2.pdf
diff --git a/Year 1/Mathematical analysis/Homeworks/Homework2.tex b/Year 1/Mathematical analysis/Homeworks/Homework2.tex
index c19a139..262e8d6 100644
--- a/Year 1/Mathematical analysis/Homeworks/Homework2.tex
+++ b/Year 1/Mathematical analysis/Homeworks/Homework2.tex
@@ -13,6 +13,7 @@
 \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
@@ -110,7 +111,7 @@
 
 \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} {{\; >0} \atop {}} \\
+	& 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
@@ -125,13 +126,13 @@
 
 \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} {{\; >0} \atop {}} \\
+	& 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} {{\; >0} \atop {}} \\
+	& 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*}
 
@@ -139,7 +140,7 @@
 
 \noindent
 \begin{align*}
-	& a_{n+1} = \frac{2a_n^2 - a_n + 6}{a_n - 6} {{\; >0} \atop {}} \implies \text{Единствения корен за } a_n \text{ е 6}
+	& 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}
@@ -156,7 +157,18 @@
 	\end{tabular}
 \end{center}
 
-[TODO: FINISH]
+\textbf{I. Случай} $\lambda < -3$ \quad Знаем, че $a_{n+1} - a_n$ е винаги намаляваща $\implies \lim\limits_{n \rightarrow \infty} a_n = -\infty$
+
+\textbf{II. Случай} $\lambda \in (-3, -2)$
+
+\textbf{III. Случай} $\lambda \in (-2, \frac{3}{2})$
+
+\textbf{IV. Случай} $\lambda \in (\frac{3}{2}, 2)$
+
+\textbf{V. Случай} $\lambda \in (2, 6)$ \quad Знаем, че $a_{n+1} - a_n$ е винаги намаляваща $\implies \lim\limits_{n \rightarrow \infty} a_n = -\infty$
+
+\textbf{VI. Случай} $\lambda > 6$ \quad Знаем, че $a_{n+1} - a_n$ е винаги растяща $\implies \lim\limits_{n \rightarrow \infty} a_n = +\infty$
+
 
 \paragraph{\colorbox{yellow}{Задача 2.}}
 
@@ -175,8 +187,6 @@
 	\qquad 	\implies L = \frac{\sqrt{2}}{\sqrt[4]{1}} = \answerbox{\sqrt{2}}
 \end{align*}
 
-\emph{2)} $\frac{3}{2}$ и 
-
 \paragraph{\colorbox{yellow}{Задача 3.}}
 
 Да се пресметне границата
@@ -187,10 +197,10 @@
 \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}{x} \quad x \rightarrow 0} + 1
+	\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}{x} \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))} \stackrel{(\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
+	\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}}
@@ -235,6 +245,20 @@
 
 \paragraph{Решение:}
 
-[TODO]
+\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 \\
+	& \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}
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