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| -rw-r--r-- | Year 1/Calculus/Homeworks/Homework3.tex | 2 |
diff --git a/Year 1/Calculus/Homeworks/Homework3.tex b/Year 1/Calculus/Homeworks/Homework3.tex index 8db531e..b8db2d3 100644 --- a/Year 1/Calculus/Homeworks/Homework3.tex +++ b/Year 1/Calculus/Homeworks/Homework3.tex @@ -62,7 +62,7 @@ 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} + & = \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.}} |