Solved 3rd exercise

2 files changed, 13 insertions, 6 deletions

diff --git a/Year 1/Mathematical analysis/Homework1.pdf b/Year 1/Mathematical analysis/Homework1.pdf
index cff64f5..fe7da74 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 1ce8387..861365b 100644
--- a/Year 1/Mathematical analysis/Homework1.tex
+++ b/Year 1/Mathematical analysis/Homework1.tex
@@ -34,7 +34,7 @@
 	f(x) = \log_{5} (x^2 - 2x + 1)
 \end{equation*}
 
-\paragraph{Решение}
+\paragraph{Решение:}
 
 [TODO]
 
@@ -48,7 +48,7 @@
 	\arcsin{\left( 2x \sqrt{1-x^2} \right)} = \pi - 2 \arccos(x)\text{, където } x \in \left[ \frac{-\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right]
 \end{equation*}
 
-\paragraph{Решение}
+\paragraph{Решение:}
 
 [TODO]
 
@@ -59,12 +59,19 @@
 Пресметнете границата на редицата
 
 \begin{equation*}
-	\lim_{x \to \infty} \frac{\sqrt[n-5]{n-5} + 2^{2n} + n^{12}}{10^n-n^3+(-1)^{3n + 5}}
+	\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}
+\paragraph{Решение:}
 
-[TODO]
+\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 \; \left( \lim\limits_{n \to \infty} (-1)^{3n + 5} < \infty \right) \\
+	& \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
 
@@ -76,7 +83,7 @@
 	\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}
+\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} = \\