Текст задатака објашњених у видео лекцији.
Одредити граничне вредности:
пр.1) $\mathop {\lim }\limits_{x \to 1} (3x - 7)$
пр.2) $\mathop {\lim }\limits_{x \to - 27} (\frac{2}{{3\sqrt[3]{x}}})$
пр.3) $\mathop {\lim }\limits_{x \to 2} (\frac{{{{(2 - x)}^3}}}{{{x^4}}})$
пр.4) $\mathop {\lim }\limits_{x \to - 3} (\frac{{2x}}{{\sqrt[3]{{9 - {x^2}}}}})$
пр.5) $\mathop {\lim }\limits_{x \to 0} (lnx)$
пр.6) $\mathop {\lim }\limits_{x \to + \infty } \left( {ln{2^x}} \right) $
пр.7) $\mathop {\lim }\limits_{x \to - \infty } \left( {ln{2^x}} \right) $
пр.1) $\mathop {\lim }\limits_{x \to 1} (3x - 7) + 3 \cdot 1 - 7 = 3 - 7 = - 4$
пр.2) $\mathop {\lim }\limits_{x \to - 27} \left( {\frac{2}{{3\sqrt[3]{x}}}} \right) = \frac{2}{{3\sqrt[3]{{ - 27}}}} = \frac{2}{{3\sqrt[3]{{{{\left( { - 3} \right)}^3}}}}} = \frac{2}{{3 \cdot \left( { - 3} \right)}} = - \frac{2}{9}$
пр.3) $\mathop {\lim }\limits_{x \to 2} \left( {\frac{{{{(2 - x)}^3}}}{{{x^4}}}} \right) = \frac{{{{(2 - 2)}^3}}}{{{2^4}}} = \frac{0}{{16}} = 0$
пр.4) $\mathop {\lim }\limits_{x \to - 3} \left( {\frac{{2x}}{{\sqrt[3]{{9 - {x^2}}}}}} \right) = \frac{{2 \cdot \left( { - 3} \right)}}{{\sqrt[3]{{9 - {{\left( { - 3} \right)}^2}}}}} = \frac{{ - 6}}{{\sqrt[3]{{9 - 9}}}} = \frac{{ - 6}}{0} = - \infty $
пр.5) $\mathop {\lim }\limits_{x \to 0} (lnx) = ln0 = - \infty $
пр.6) $\mathop {\lim }\limits_{x \to + \infty } \left( {ln{2^x}} \right) = + \infty $
пр.7) $\mathop {\lim }\limits_{x \to - \infty } \left( {ln{2^x}} \right) = 0 $
