جواب کاردرکلاس صفحه 53 درس 3 ریاضی دوازدهم تجربی (حد بی نهایت و حد در بی نهایت)

الف \(\mathop {\lim }\limits_{x \to - 3} \;\frac{{{x^2} - 9}}{{{x^2} + 3x}}\)

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

ب \(\mathop {\lim }\limits_{x \to \frac{1}{2}} \;\frac{{4{x^2} - 4x + 1}}{{2{x^2} + x - 1}}\)

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

پ \(\mathop {\lim }\limits_{x \to \frac{1}{2}} \;\frac{{4{x^2} - 1}}{{2{x^3} - 13{x^2} + 24x - 9}}\)

\( = \mathop {\lim }\limits_{x \to \;\frac{1}{2}} \frac{{(2x - 1)(2x + 1)}}{{(2x - 1)({x^2} - 6x + 9)}} = \mathop {\lim }\limits_{x \to \;\frac{1}{2}} \frac{{2x + 1}}{{{{(x - 3)}^2}}} = \frac{{\frac{5}{2}}}{{\frac{{25}}{4}}} = \frac{2}{5}\)

ت \(\mathop {\lim }\limits_{x \to - 1} \;\frac{{{x^2} - 1}}{{x + \sqrt {2x + 3} }}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to - 1} \;\frac{{{x^2} - 1}}{{x + \sqrt {2x + 3} }} \times \frac{{x - \sqrt {2x + 3} }}{{x - \sqrt {2x + 3} }} = \mathop {\lim }\limits_{x \to \; - \;1} \frac{{({x^2} - 1)(x - \sqrt {2x + 3} )}}{{{x^2} - 2x - 3}}\\\\ = \mathop {\lim }\limits_{x \to \; - \;1} \frac{{(x - 1)(x + 1)(x - \sqrt {2x + 3} )}}{{(x + 1)(x - 3)}} = \mathop {\lim }\limits_{x \to \; - \;1} \frac{{(x - 1)(x - \sqrt {2x + 3} )}}{{x - 3}} = - 1\end{array}\)

ث \(\mathop {\lim }\limits_{x \to 1} \;\frac{{x - \sqrt x }}{{{x^2} + x - 2}}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \;\frac{{x - \sqrt x }}{{{x^2} + x - 2}} \times \frac{{x + \sqrt x }}{{x + \sqrt x }} = \mathop {\lim }\limits_{x \to \;1} \frac{{{x^2} - x}}{{(x - 1)(x + 2)(x + \sqrt x )}}\\\\ = \mathop {\lim }\limits_{x \to \;1} \frac{{x(x - 1)}}{{(x - 1)(x + 2)(x + \sqrt x )}} = \mathop {\lim }\limits_{x \to \;1} \frac{x}{{(x + 2)(x + \sqrt x )}} = \frac{1}{{3 \times 2}} = \frac{1}{6}\end{array}\)

ج \(\mathop {\lim }\limits_{x \to - 1} \;\frac{{\sqrt[3]{x} + 1}}{{{x^2} + 3x + 2}}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to - 1} \;\frac{{\sqrt[3]{x} + 1}}{{{x^2} + 3x + 2}} \times \frac{{\sqrt[3]{{{x^2}}} - \sqrt[3]{x} + 1}}{{\sqrt[3]{{{x^2}}} - \sqrt[3]{x} + 1}} = \mathop {\lim }\limits_{x \to \; - \;1} \frac{{x + 1}}{{(x + 1)(x + 2)(\sqrt[3]{{{x^2}}} - \sqrt[3]{x} + 1)}}\\\\ = \mathop {\lim }\limits_{x \to \; - \;1} \frac{1}{{(x + 2)(\sqrt[3]{{{x^2}}} - \sqrt[3]{x} + 1)}} = \frac{1}{{1 \times 1}} = 1\end{array}\)