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حدود زیر را محاسبه کنید:
الف \(\mathop {\lim }\limits_{x \to + \infty } \;\frac{{2{x^3} - 5x + 4}}{{7{x^3} - 11{x^2} - 6x}}\)
ب \(\mathop {\lim }\limits_{x \to - \infty } \;\frac{{5x + 4}}{{{x^3} + x - 8}}\)
پ \(\mathop {\lim }\limits_{x \to - \infty } \;\frac{{ - 4{x^7} + 5{x^2}}}{{2{x^3} + 9}}\)
الف
\(\mathop {\lim }\limits_{x \to + \infty } \;\frac{{2{x^3} - 5x + 4}}{{7{x^3} - 11{x^2} - 6x}} = \mathop {\lim }\limits_{x \to \; + \;\infty } \frac{{2{x^3}}}{{7{x^3}}} = \frac{2}{7}\)
ب
\(\mathop {\lim }\limits_{x \to - \infty } \;\frac{{5x + 4}}{{{x^3} + x - 8}} = \mathop {\lim }\limits_{x \to \; - \;\infty } \frac{{5x}}{{{x^3}}} = \mathop {\lim }\limits_{x \to \; - \;\infty } \frac{5}{{{x^2}}} = 0\)
پ
\(\mathop {\lim }\limits_{x \to - \infty } \;\frac{{ - 4{x^7} + 5{x^2}}}{{2{x^3} + 9}} = \mathop {\lim }\limits_{x \to \; - \;\infty } \frac{{ - 4{x^7}}}{{2{x^3}}} = \mathop {\lim }\limits_{x \to \; - \;\infty } \frac{{ - 4{x^4}}}{2} = - \;\infty \)
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