جواب تمرین صفحه 127 درس 6 ریاضی یازدهم تجربی (حد و پیوستگی)

الف نادرست

ب درست

پ نادرست

ت درست

ث نادرست

ج درست

چ درست

ح درست

الف

\(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \sqrt {x - 2} = \sqrt {{2^ + } - 2} = \sqrt {{0^ + }} = 0\)

ب

\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \sqrt {x - 2} = \sqrt {{2^ - } - 2} = \sqrt {{0^ - }} = \not \exists \)

پ

\(\begin{array}{l}\begin{array}{*{20}{l}}\begin{array}{l}\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 0\\\end{array}\\{\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \not \exists }\end{array}\\\\ \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)\\\\ \Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \not \exists \end{array}\)

ت

\(f\left( 2 \right) = \sqrt {2 - 2} = \sqrt 0 = 0\)

بله؛ حد دارد:

\(\begin{array}{l}\begin{array}{*{20}{l}}\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} x = 0\\\end{array}\\{\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - x} \right) = 0}\end{array}\\\\ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = 0\\\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} f\left( x \right) = 0\end{array}\)

خیر؛ f(0) تعریف نشده است.

الف

f(2) = 2(2) + 1 = 5

g(2) = وجود ندارد

h(2) = 3

ب

\(\begin{array}{*{20}{l}}\begin{array}{l}\mathop {\lim }\limits_{x \to 2} h\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left( {2x + 1} \right) = 2\left( 2 \right) + 1 = 5\\\end{array}\\\begin{array}{l}\mathop {\lim }\limits_{x \to 2} g\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left( {2x + 1} \right) = 2\left( 2 \right) + 1 = 5\\\end{array}\\{\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left( {2x + 1} \right) = 2\left( 2 \right) + 1 = 5}\end{array}\)

\(\begin{array}{l}\begin{array}{*{20}{l}}\begin{array}{l}\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( { - x + 2} \right) = - 2 + 2 = 0\\\end{array}\\{\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {x - 3} \right) = 2 - 3 = - 1}\end{array}\\\\ \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)\\\\ \Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \not \exists \end{array}\)

وجود ندارد

\(\begin{array}{l}\begin{array}{*{20}{l}}\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {{x^2} + 2} \right) = \\\\{\left( 0 \right)^2} + 2 = 2\\\end{array}\\\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - 2x - 2} \right) = \\\\ - 2\left( 0 \right) - 2 = - 2\end{array}\end{array}\\\\ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\\\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \not \exists \end{array}\)

 وجود ندارد

\(\begin{array}{l}f\left( x \right) = \frac{{\left| x \right|}}{x} = \left\{ {\begin{array}{*{20}{l}}{1\;\;\:,\;\;\:\;\;\:x > 0}\\{ - 1\;\;\:,\;\;\:x < 0}\end{array}} \right.\\\\\left. {\begin{array}{*{20}{l}}\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( 1 \right) = 1\\\end{array}\\{\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - 1} \right) = - 1}\end{array}} \right\}\\\\ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\\\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \not \exists \end{array}\)

وجود ندارد