جواب فعالیت صفحه 131 درس 6 ریاضی یازدهم تجربی (حد و پیوستگی)

الف

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \;f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \;\left( {3{x^2} + 2x - 7} \right)\\\\ = \mathop {\lim }\limits_{x \to 1} \;\left( {3{x^2}} \right) + \mathop {\lim }\limits_{x \to 1} \;\left( {2x} \right) - \mathop {\lim }\limits_{x \to 1} 7\\\\ = 3\mathop {\lim }\limits_{x \to 1} \;{x^2} + \;2\mathop {\lim }\limits_{x \to 1} x + \left( { - \mathop {\lim }\limits_{x \to 1} 7} \right)\\\\ = 3{\left( 1 \right)^2} + 2\left( 1 \right) - 7 = - 2\end{array}\)

ب

\(\begin{array}{l}\left\{ \begin{array}{l}f\left( 1 \right) = 3{\left( 1 \right)^2} + 2\left( 1 \right) - 7 = - 2\\\\\mathop {\lim }\limits_{x \to 1} f\left( x \right) = - 2\end{array} \right.\\\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)\end{array}\)

پ

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 2} g\left( x \right) = \mathop {\lim }\limits_{x \to 2} (\frac{1}{8}{x^4} - {x^3} + 5x - \frac{1}{2}) = \\\\\mathop {\lim }\limits_{x \to 2} \frac{1}{8}{x^4} - \mathop {\lim }\limits_{x \to 2} {x^3} + \mathop {\lim }\limits_{x \to 2} 5x - \mathop {\lim }\limits_{x \to 2} \frac{1}{2} = \\\\ = \frac{1}{8}\mathop {\lim }\limits_{x \to 2} {x^4} - \mathop {\lim }\limits_{x \to 2} {x^3} + \mathop {5\lim }\limits_{x \to 2} x - \mathop {\lim }\limits_{x \to 2} \frac{1}{2} = \\\\\frac{1}{8}{\left( 2 \right)^4} - {\left( 2 \right)^3} + 5\left( 2 \right) - \frac{1}{2} = \frac{7}{2}\\\\g\left( 2 \right) = \frac{1}{8}{\left( 2 \right)^4} - {\left( 2 \right)^3} + 5\left( 2 \right) - \frac{1}{2} = \frac{7}{2}\\\\\mathop { \Rightarrow \lim }\limits_{x \to 2} g\left( x \right) = g\left( 2 \right)\end{array}\)

الف

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \;\frac{{2x - 1}}{{{x^2} - 4x + 1}} = \;\\\\\frac{{\mathop {\lim }\limits_{x \to 3} \left( {2x - 1} \right)}}{{\mathop {\lim }\limits_{x \to 3} \left( {{x^2} - 4x + 1} \right)}} = \\\\\frac{{\mathop {\lim }\limits_{x \to 3} 2x - \mathop {\lim }\limits_{x \to 3} 1}}{{\mathop {\lim }\limits_{x \to 3} {x^2} - \mathop {\lim }\limits_{x \to 3} 4x + \mathop {\lim }\limits_{x \to 3} 1}} = \\\\\frac{{2\mathop {\lim }\limits_{x \to 3} x - \mathop {\lim }\limits_{x \to 3} 1}}{{\mathop {\lim }\limits_{x \to 3} {x^2} - 4\mathop {\lim }\limits_{x \to 3} x + \mathop {\lim }\limits_{x \to 3} 1}} = \\\\\frac{{2\left( 3 \right) - 1}}{{{{\left( 2 \right)}^2} - 4\left( 2 \right) + 1}} = \frac{5}{{ - 2}} = - 2/5\end{array}\)

ب

\(\begin{array}{l}\mathop {\lim }\limits_{x \to } \;\frac{{{x^4} + 2{x^3} + 1}}{{5{x^2} + \frac{2}{3}}} = \\\\\frac{{\mathop {\lim }\limits_{x \to 1} \left( {{x^4} + 2{x^3} + 1} \right)}}{{\mathop {\lim }\limits_{x \to 1} (5{x^2} + \frac{2}{3})}} = \\\\\frac{{\mathop {\lim }\limits_{x \to 1} {x^4} + 2\mathop {\lim }\limits_{x \to 1} {x^3} + \mathop {\lim }\limits_{x \to 1} 1}}{{5\mathop {\lim }\limits_{x \to 1} {x^2} + \mathop {\lim }\limits_{x \to 1} \frac{2}{3}}} = \\\\\frac{{{{\left( 1 \right)}^4} + 2{{\left( 1 \right)}^3} + 1}}{{5{{\left( 1 \right)}^2} + \frac{2}{3}}} = \frac{{12}}{{17}}\end{array}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to } \;\frac{{{x^2} - 1}}{{\frac{3}{5}{x^2} - 2x + 1}} = \\\\\frac{{\mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right)}}{{\mathop {\lim }\limits_{x \to 1} (\frac{3}{5}{x^2} - 2x + 1)}} = \\\\\frac{{\mathop {\lim }\limits_{x \to 1} {x^2} - \mathop {\lim }\limits_{x \to 1} 1}}{{\frac{3}{5}\mathop {\lim }\limits_{x \to 1} {x^2} - 2\mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 1}} = \\\\\frac{{{{\left( 1 \right)}^2} - 1}}{{\frac{3}{5}{{\left( 1 \right)}^2} - 2\left( 1 \right) + 1}} = 0\end{array}\)