جواب کاردرکلاس صفحه 55 درس 3 حسابان دوازدهم (حدهای نامتناهی_حد در بی نهایت)

اگر \(\mathop {\lim }\limits_{x \to \;a} f(x) = - \;\infty \)  و \(\mathop {\lim }\limits_{x \to \;a} g(x) = L\)  آنگاه \(\mathop {\lim }\limits_{x \to \;a} (f + g)(x) = - \;\infty \)

اگر \(\mathop {\lim }\limits_{x \to \;a} f(x) = - \;\infty \) و \(\mathop {\lim }\limits_{x \to \;a} g(x) = L\) آنگاه اگر L > 0  در نتیجه \(\mathop {\lim }\limits_{x \to \;a} (f \times g)(x) = - \;\infty \) و اگر L < 0 در نتیجه \(\mathop {\lim }\limits_{x \to \;a} (f \times g)(x) = + \;\infty \)

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

ب \(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^2} + x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to \;{0^ + }} \frac{{x(x + 1)}}{{{x^2}}} = \)

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

پ \(\mathop {\lim }\limits_{x \to - {2^ + }} \frac{{x + 2}}{{{x^2} + 4x + 4}} = \mathop {\lim }\limits_{x \to \; - {2^ + }} \frac{{x + 2}}{{{{(x + 2)}^2}}} = \)

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

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