جواب تمرین صفحه 135 درس 6 ریاضی یازدهم تجربی (حد و پیوستگی)

الف \(\mathop {\lim }\limits_{x \to 3} \;f\left( x \right) = 2\)

ب \(\mathop {\lim }\limits_{x \to - 1} \;f\left( x \right) = \not \exists \) وجود ندارد

پ \(\mathop {\lim }\limits_{x \to 3} \;g\left( x \right) = {\rm{ }}5\)

ت \(\mathop {\lim }\limits_{x \to 3} \;\left( {f\left( x \right) + g\left( x \right)} \right) = {\rm{ }}\left( {2{\rm{ }} + {\rm{ }}5} \right){\rm{ }} = {\rm{ }}7\)

ث \(\mathop {\lim }\limits_{x \to - 1} \;\left( {f\left( x \right) + g\left( x \right)} \right) = \not \exists \) وجود ندارد

ج \(\mathop {\lim }\limits_{x \to 2} \;\left( {2f\left( x \right) + 5g\left( x \right)} \right) = \not \exists \) وجود ندارد

چ \(\mathop {\lim }\limits_{x \to 0} \;{\left( {f\left( x \right)} \right)^4} = {(\frac{1}{3})^4} = \frac{1}{{81}}\)

ح \(\mathop {\lim }\limits_{x \to 0} \;{\left( {g\left( x \right)} \right)^2} = \not \exists \) وجود ندارد

خ \(\mathop {\lim }\limits_{x \to 2} \;\frac{{f\left( x \right)}}{{g\left( x \right)}} = \not \exists \) وجود ندارد

د \(\mathop {\lim }\limits_{x \to 5} \left( {f\left( x \right)\,.\,g\left( x \right)} \right) = {\rm{ }}(5 \times 5){\rm{ }} = {\rm{ }}25\)

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{l}}\begin{array}{l}f\left( x \right) = \sqrt x \\\end{array}\\{g\left( x \right) = \left| x \right|}\end{array}} \right.\\\\\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \sqrt 1 = 1\\\\\mathop {\lim }\limits_{x \to 1} g\left( x \right) = \left| 1 \right| = 1\end{array} \right.\\\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} g\left( x \right)\end{array}\)

الف \(\mathop {\lim }\limits_{x \to 7} \;\left( { - 3} \right) = - 3\)

ب \(\mathop {\lim }\limits_{x \to 0} \;\left( { - 2x - 7} \right) = - 2\left( 0 \right) - 7 = - 7\)

پ \(\mathop {\lim }\limits_{x \to - 1} \;\left( {3{x^2} - 4x + 5} \right) = \)

\(3{\left( { - 1} \right)^2} - 4\left( { - 1} \right) + 5 = 12\)

ت \(\mathop {\lim }\limits_{x \to 3} \;\frac{{{x^2} - 3x}}{{{x^2} - 9}} = \mathop {\lim }\limits_{x \to 3} \,\,\frac{{x\left( {x - 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \)

\(\mathop {\lim }\limits_{x \to 3} \,\,\frac{x}{{\left( {x + 3} \right)}} = \frac{3}{{3 + 3}} = \frac{1}{2}\)

ث \(\mathop {\lim }\limits_{x \to 0} \;\frac{x}{{2{x^2} - x}} = \mathop {\lim }\limits_{x \to 0} \;\frac{x}{{x\left( {2x - 1} \right)}} = \)

\(\mathop {\lim }\limits_{x \to 0} \;\frac{1}{{2x - 1}} = \frac{1}{{2\left( 0 \right) - 1}} = - 1\)

ج \(\mathop {\lim }\limits_{x \to - 2} \;\frac{{{x^3} + 8}}{{x + 2}} = \)

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

چ \(\mathop {\lim }\limits_{x \to - 2} \;\left[ x \right] = \not \exists \)وجود ندارد 

ح \(\mathop {\lim }\limits_{x \to {0^ + }} \;\sqrt x = \sqrt {{0^ + }} = 0\)

خ \(\mathop {\lim }\limits_{x \to 2} \;\sqrt {x + 7} = \sqrt {2 + 7} = \sqrt 9 = 3\)

د \(\mathop {\lim }\limits_{x \to {0^ - }} \;\sqrt x = \sqrt {{0^ - }} = \not \exists \) وجود ندارد

ذ \(\mathop {\lim }\limits_{x \to 2} \;\sqrt {x + 5} = \sqrt {2 + 5} = \sqrt 7 \)

ر \(\mathop {\lim }\limits_{x \to 1} \;\sqrt {x - 2} = \sqrt {1 - 2} = \sqrt { - 1} = \not \exists \) وجود ندارد

ز \(\mathop {\lim }\limits_{x \to {3^ + }} \;\frac{{x - 2}}{{\left[ x \right] + 1}} = \frac{{{3^ + } - 2}}{{\left[ {{3^ + }} \right] + 1}} = \frac{{{1^ + }}}{{3 + 1}} = \frac{1}{4}\)

ژ \(\mathop {\lim }\limits_{x \to - \frac{\pi }{3}} \;\cos x = \cos ( - \frac{\pi }{3}) = \cos \frac{\pi }{3} = \frac{1}{2}\)

س \(\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \;\left( {\sin x + \cos x} \right) = \)

\(\sin \frac{\pi }{4} + \cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} = \sqrt 2 \)

ش \(\mathop {\lim }\limits_{x \to - {2^ + }} \;\frac{x}{{\left[ x \right]}} = \frac{{ - {2^ + }}}{{\left[ { - {2^ + }} \right]}} = \frac{{ - 2}}{{ - 2}} = 1\)

ص \(\mathop {\lim }\limits_{x \to 2} \;\frac{{{x^2} + x - 6}}{{{x^2} - 6x + 8}} = \)

\(\begin{array}{l} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}}{{\left( {1 - \sin x} \right)}} = \\\\\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \left( {1 + \sin x} \right) = 1 + \sin \frac{\pi }{2} = \\\\1 + 1 = 2\end{array}\)

ض \(\mathop {\lim }\limits_{x \to 0} \;\left( {x + \left[ x \right]} \right) = \not \exists \) وجود ندارد

الف \(\mathop {\lim }\limits_{x \to 2} \;\left( {f\left( x \right) + h\left( x \right)} \right) = \)

\(\mathop {\lim }\limits_{x \to 2} f\left( x \right) + \mathop {\lim }\limits_{x \to 2} h\left( x \right) = 3 + \left( { - 1} \right) = 2\)

ب \(\mathop {\lim }\limits_{x \to 2} \;{\left( {h\left( x \right)} \right)^5} = {\left( { - 1} \right)^5} = - 1\)

پ \(\mathop {\lim }\limits_{x \to 2} \;\frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to 2} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 2} g\left( x \right)}} = \frac{3}{0} = \not \exists \) وجود ندارد

ت \(\mathop {\lim }\limits_{x \to 2} \;\frac{{g\left( x \right)}}{{f\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to 2} g\left( x \right)}}{{\mathop {\lim }\limits_{x \to 2} f\left( x \right)}} = \frac{0}{3} = 0\)

ث \(\mathop {\lim }\limits_{x \to 2} \;\frac{{3f\left( x \right)}}{{g\left( x \right) - 5h\left( x \right)}} = \)

\(\frac{{3\mathop {\lim }\limits_{x \to 2} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 2} g\left( x \right) - 5\mathop {\lim }\limits_{x \to 2} h\left( x \right)}} = \frac{{3\left( 3 \right)}}{{\left( 0 \right) - 5\left( { - 1} \right)}} = \frac{9}{5}\)

ج \(\mathop {\lim }\limits_{x \to 2} \;\frac{1}{{h\left( x \right)}} = \frac{1}{{\mathop {\lim }\limits_{x \to 2} h\left( x \right)}} = \frac{1}{{ - 1}} = - 1\)

\(\begin{array}{l}f\left( x \right) = \frac{{\left| {x - 3} \right|}}{{x - 3}} = \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}1\;\;\:\:\:{\mkern 1mu} ,\;\;\:x > 3\\\end{array}\\{ - 1\;\;\:,\;\;\:x < 3}\end{array}} \right.\\\\g\left( x \right) = 1\\\\\left\{ {\begin{array}{*{20}{l}}\begin{array}{l}\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = 1\\\end{array}\\{\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = - 1}\end{array}} \right.\\\\ \Rightarrow \mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right)\end{array}\)

\( \Rightarrow \mathop {\lim }\limits_{x \to 3} f\left( x \right) = \not \exists \) وجود ندارد

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \,g(x) = 1\\\\{x_0} > 3 \Rightarrow \mathop {\lim }\limits_{x \to {x_0}} \,f(x) = \mathop {\lim }\limits_{x \to {x_0}} \,g(x)\\\\ \Rightarrow {x_0} \in (3\,,\,\infty )\end{array}\)