جواب کاردرکلاس صفحه 60 درس 3 ریاضی دوازدهم تجربی (حد بی نهایت و حد در بی نهایت)

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

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

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

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

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

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

الف

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

ب

\(\begin{array}{l}g(x) = \frac{{200x + 300}}{{2x - 5}}\\\\ \Rightarrow \mathop {\lim }\limits_{x \to \; - \;\infty } g(x) = \mathop {\lim }\limits_{x \to \; - \;\infty } \frac{{200x + 300}}{{2x - 5}} = \frac{{200}}{2} = 100\end{array}\)