Solved ex1 from dis hw3 and added answerbox to ex5

2 files changed, 14 insertions, 2 deletions

diff --git a/Year 1/Calculus/Homeworks/Homework3.pdf b/Year 1/Calculus/Homeworks/Homework3.pdf
index aa2ee84..f69c9ae 100644
--- a/Year 1/Calculus/Homeworks/Homework3.pdf
+++ b/Year 1/Calculus/Homeworks/Homework3.pdf
diff --git a/Year 1/Calculus/Homeworks/Homework3.tex b/Year 1/Calculus/Homeworks/Homework3.tex
index 69ad5f1..8db531e 100644
--- a/Year 1/Calculus/Homeworks/Homework3.tex
+++ b/Year 1/Calculus/Homeworks/Homework3.tex
@@ -51,7 +51,19 @@
 	\lim_{x \rightarrow 0} \frac{\ln (x^2+1) - 3e^x + \sqrt[4]{12x + 1} + 2}{\tg(x^2) - 1 + \cos x}.
 \end{align*}
 
-\paragraph{Решение:} [TODO]
+\paragraph{Решение:} Развиваме функциите в дробта по Маклорен:
+\begin{align*}
+	& \ln (x^2 + 1) = x^2 + o(x^2) \qquad e^x = 1 + x + \frac{x^2}{2} + o(x^2) \qquad \cos x = 1 - \frac{x^2}{2} + o(x^2) \\
+	& (1 + 12x)^{\frac{1}{4}} = 1 + \frac{1}{4}12x + \binom{\frac{1}{4}}{2}x^2 + o(x^2) = 1 + 3x - \frac{27}{2}x^2 + o(x^2) \\
+	& \quad (\tg (x^2))' = \frac{(\sin (x^2))' \cos (x^2) - \sin (x^2) (\cos (x^2))'}{\cos^2 (x^2)} = \frac{2x \cos^2(x^2) + 2x \sin^2(x^2)}{\cos ^2 (x^2)} = \frac{2x}{\cos ^2 (x^2)} \\
+	& \quad (\tg (x^2))'' = \left( \frac{2x}{\cos ^2 (x^2)} \right)' = \frac{(2x)'\cos^2 (x^2) - 2x(\cos^2 (x^2))'}{\cos^4(x^2)} = \frac{2\cos^2 (x^2) + 8x^2\sin (x^2) \cos(x^2)}{\cos^4(x^2)} \\
+	& \quad \tg(0) = 0, \quad (\tg(x^2))'(0) = 0, \quad (\tg(x^2))''(0) = \frac{2}{2} x^2 \implies \tg (x^2) = x^2 + o(x^2)
+\end{align*}
+Тоест границата е еквивалентна на:
+\begin{align*}
+	& \lim_{x \rightarrow 0} \frac{x^2 + -3 \left( 1 + x + \frac{x^2}{2} \right) + 1 + 3x - \frac{27}{2}x^2 + 2 + o(x^2)}{x^2 - 1 + 1 - \frac{x^2}{2} + o(x^2)} = \\
+	& \lim_{x \rightarrow 0} \frac{x^2 \cancel{- 3} \cancel{- 3x} -3 \frac{x^2}{2} + \cancel{1} + \cancel{3x} - 27 \frac{x^2}{2} + \cancel{2} + o(x^2)}{\frac{x^2}{2} + o(x^2)} = \lim_{x \rightarrow 0} \frac{-\frac{28}{2}x^2 + o(x^2)}{\frac{x^2}{2} + o(x^2)} = \lim_{x \rightarrow 0} \frac{x^2 \left( -\frac{28}{2} + \frac{o(x^2)}{x^2} \right)}{x^2 \left( \frac{1}{2} + \frac{o(x^2)}{x^2} \right)} = \answerbox{-28}
+\end{align*}
 
 \paragraph{\colorbox{yellow}{Задача 2.}}
 
@@ -136,7 +148,7 @@
 \begin{align*}
 	& \int x \arctg(x^2) \D x = \int \arctg(x^2) \D \left( \frac{x^2}{2} \right) \xlongequal{\text{По части}} \frac{x^2}{2}\arctg(x^2) - \int \frac{x^2}{2} \D \arctg(x^2) \\
 	& \qquad \int \frac{x^2}{2} \D \arctg(x^2) = \int \frac{x^2}{\cancel{2}} \frac{1}{x^4+1} \cancel{2}x \D x = \int \frac{1}{x^4+1} \cdot \frac{4}{4} x^3 \D x = \frac{1}{4} \int \frac{1}{x^4+1} d(x^4) = \frac{1}{4} \ln (x^4 + 1) + c \\
-	& \implies \int x \arctg(x^2) \D x = \frac{x^2}{2}\arctg(x^2) - \frac{1}{4} \ln (x^4 + 1) + c
+	& \implies \int x \arctg(x^2) \D x = \answerbox{\frac{x^2}{2}\arctg(x^2) - \frac{1}{4} \ln (x^4 + 1) + c}
 \end{align*}
 
 \end{document}
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