جواب تمرین صفحه 63 درس 3 ریاضی دوازدهم تجربی (حد بی نهایت و حد در بی نهایت)

الف

\(f\left( x \right) = \frac{1}{x}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \;f\left( x \right) = \;0\\\\\mathop {\lim }\limits_{x \to + \infty } \;f\left( x \right) = \;0\end{array}\)

\(\mathop {\lim }\limits_{x \to 0} \;f\left( x \right) = \;\)وجود ندارد

ب

\(g\left( x \right) = \left\{ \begin{array}{l}1\;\;\;\;\;\;\;\;x > 0\\ - 1\;\;\;\;\;\;x < 0\end{array} \right.\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \;g\left( x \right) = 1\\\\\mathop {\lim }\limits_{x \to + \infty } \;g\left( x \right) = - \;1\end{array}\)

الف 

\(\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \;f\left( x \right) = 2\\\\\mathop {\lim }\limits_{x \to + \infty } \;f\left( x \right) = 0\end{array}\)

ب 

\(\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \;g\left( x \right) = - \;\infty \\\\\mathop {\lim }\limits_{x \to + \infty } \;g\left( x \right) = 2\end{array}\)

پ 

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

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

ب \(\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \;f\left( x \right) = + \;\infty \)

پ \(\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \;f\left( x \right) = - \;\infty \)

ت \(\mathop {\lim }\limits_{x \to {1^ - }} \;f\left( x \right) = + \;\infty \)

ث \(\mathop {\lim }\limits_{x \to {1^ + }} \;f\left( x \right) = + \;\infty \)

ج \(\mathop {\lim }\limits_{x \to + \infty } \;f\left( x \right) = 1\)

الف 

\(\mathop {\lim }\limits_{x \to - \infty } \;(9 + \frac{7}{{{x^3}}}) = 9 + 0 = 9\)

ب 

\(\mathop {\lim }\limits_{x \to + \infty } \;( - \frac{1}{2}{x^3} + 7{x^2} - 6) = \mathop {\lim }\limits_{x \to \; + \;\infty } ( - \frac{1}{2}{x^3}) = - \;\infty \)

پ 

\(\mathop {\lim }\limits_{x \to - \infty } \;\frac{1}{{2x - 3}} = \mathop {\lim }\limits_{x \to \; - \;\infty } (\frac{1}{{2x}}) = 0\)

ت 

\(\mathop {\lim }\limits_{x \to + \infty } \;\frac{{3 + \frac{1}{{{x^2}}}}}{{\frac{4}{x} - 5}} = \frac{{3 + \mathop {\lim }\limits_{x \to \; + \;\infty } \frac{1}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to \; + \;\infty } \frac{4}{x} - 5}} = \frac{{3 + 0}}{{0 - 5}} = - \;\frac{3}{5}\)

ث 

\(\mathop {\lim }\limits_{x \to + \infty } \;\frac{{2x - 1}}{{3x + 1}} = \mathop {\lim }\limits_{x \to \; + \;\infty } (\frac{{2x}}{{3x}}) = \frac{2}{3}\)

ج 

\(\mathop {\lim }\limits_{x \to - \infty } \;\frac{{2{x^2} - 3x + 1}}{{{x^2} + 5x - 3}} = \mathop {\lim }\limits_{x \to \; - \;\infty } \frac{{2{x^2}}}{{{x^2}}} = 2\)

چ 

\(\mathop {\lim }\limits_{x \to - \infty } \;\frac{{2{x^5} - 6{x^3} - x}}{{{x^2} - 5x + 1}} = \mathop {\lim }\limits_{x \to \; - \;\infty } (\frac{{2{x^5}}}{{{x^2}}}) = \mathop {\lim }\limits_{x \to \; - \;\infty } 2{x^3} = - \;\infty \)

ح 

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

خ 

\(\mathop {\lim }\limits_{x \to - \infty } \;\frac{{ - 6{x^3} + 7x - 9}}{{2{x^3} - 4{x^2} + x}} = \mathop {\lim }\limits_{x \to \; - \;\infty } (\frac{{ - 6{x^3}}}{{2{x^3}}}) = - \;3\)

د 

\(\mathop {\lim }\limits_{x \to + \infty } \;\frac{{2x + 1}}{4} = \mathop {\lim }\limits_{x \to \; + \;\infty } \frac{{2x}}{4} = + \;\infty \)

الف در مورد عبارت \(\mathop {\lim }\limits_{x \to \; + \;\infty } f(x) = \; - \;1\) اگر x به اندازه کافی بزرگ انتخاب شود، تابع f(x) را به هر اندازه دلخواه می توان به 1- نزدیکتر کرد.

در مورد عبارت \(\mathop {\lim }\limits_{x \to \; - \infty } f(x) = \;2\) اگر x به اندازه کافی کوچک انتخاب شود، تابع f(x) را به هر اندازه دلخواه می توان به 2 نزدیکتر کرد.

ب

مسئله بیشمار جواب دارد.