نمونه سوالات فصل 3 حسابان دوازدهم همراه با جواب تشریحی

\(\frac{2}{3}\)

\(\begin{array}{l} - \infty \\\\ + \infty \\\\ - 2\end{array}\)

\(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{{x^2} - 1}} \times \frac{{\sqrt x + 1}}{{\sqrt x + 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{x - 1}}{{(x - 1)(x + 1)\sqrt x + 1}}\frac{1}{4}\)

\(\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}} \times \frac{{\sqrt x + 3}}{{\sqrt x + 3}} = \mathop {\lim }\limits_{x \to 9} \frac{{x - 9}}{{(x - 9)(\sqrt {x + 3} )}} = \frac{1}{6}\)

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

\(\mathop {\lim }\limits_{x \to - \infty } \frac{{ - x}}{{5x}} = - \frac{1}{5}\)

4

\(\begin{array}{l}1)\,\mathop {\lim }\limits_{x \to - \infty } f(x) = 2\\\\2)\,\mathop {\lim }\limits_{x \to + \infty } f(x) = 0\\\\3)\,\mathop {\lim }\limits_{x \to - 1} f(x) = + \infty \\\\4)\,\mathop {\lim }\limits_{x \to {1^ - }} f(x) = - \infty \end{array}\)

\(\mathop {\lim }\limits_{x \to 3} \frac{{(x - 3)(x - 2)}}{{(x - 3)(2x - 1)}} = \frac{1}{5}\)

\((x + 1)\)

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

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

\(\mathop {\lim }\limits_{x \to 5} \frac{{(2 - \sqrt {x - 1} )}}{{(x - 5)(2 + \sqrt {x - 1} )}} = \mathop {\lim }\limits_{x \to 5} \frac{{ - (x - 5)}}{{(x - 5)(2 + \sqrt {x - 1} )}} = \frac{{ - 1}}{4}\)

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

\(\frac{{3 + 0}}{{0 - 5}} = - \frac{3}{5}\)

\(\begin{array}{l}f(x) = {x^3} - a{x^2} + 5x - b\\\\\left\{ \begin{array}{l}f(1) = 1 - a - b = 0\\\\f( - 3) = - 27 - 9a - 15 - b = 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a + b = 6\\\\ - 9a - b = + 42\end{array} \right.\\\\ \Rightarrow - 8a = 48 \Rightarrow a = - 6\,,\,b = 12\end{array}\)

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

\(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{[x] - 3}}{{{{\sin }^3}x}} = \frac{{[{0^ + }] - 3}}{{{{\sin }^3}({0^ + })}} = \frac{{0 - 3}}{{{0^ + }}} = - \infty \)

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

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

\({x^2} - ax + b = {(x - 4)^2} = {x^2} + 16 - 8x \Rightarrow a = + 8\,,\,b = 16\)

\(\begin{array}{l}1)\,\mathop {\lim }\limits_{x \to - {4^ + }} f(x) = - \infty \\\\2)\,\mathop {\lim }\limits_{x \to {3^ + }} f(x) = - \infty \\\\3)\,\mathop {\lim }\limits_{x \to {3^ - }} f(x) = + \infty \\\\4)\,\mathop {\lim }\limits_{x \to - {2^ - }} f(x0 = 2\\\\5)\,\mathop {\lim }\limits_{x \to 1} f(x) = \Rightarrow \left\{ \begin{array}{l} = 3\\\\ = - 1\end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to 1} f(x) \otimes \end{array}\)

\(\mathop {\lim }\limits_{x \to + \infty } \frac{{5 - \frac{1}{{{x^2}}}}}{{\frac{8}{x} - 3}} = \frac{{5 - 0}}{{0 - 3}} = - \frac{5}{3}\)

\(\mathop {\lim }\limits_{x \to - \infty } \frac{{ - 7{x^4} + 2{x^2} - 3}}{{16{x^3} - 3}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 7x}}{{16}} = ( - \frac{7}{{16}}) \times ( - \infty ) = + \infty \)

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

\(\mathop {\lim }\limits_{x \to - \infty } - 17 + \frac{3}{{ - {x^3} + \sqrt x }} = - 17 + \frac{3}{{ - {{( - \infty )}^3}}} = - 17 + 0 = - 17\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } f(x) = \mathop {\lim }\limits_{x \to + \infty } \frac{1}{x} = 0\\\\\mathop {\lim }\limits_{x \to - \infty } f(x) = \mathop {\lim }\limits_{x \to - \infty } \frac{{3{x^2} - 5x}}{{ - {x^2} + 1}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{3{x^2}}}{{ - {x^2}}} = - 3\end{array}\)

\(\mathop {\lim }\limits_{x \to - \infty } f(x) = \mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^3} + 3{x^2} - x}}{{5{x^3} - 2{x^5} + 4}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^3}}}{{ - 2{x^5}}} = 0\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } f(x) = - \infty \\\\\mathop {\lim }\limits_{x \to - \infty } f(x) = + 2\end{array}\)

\( - \infty \)

\(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{[x]}}{{\sin x}} = \frac{{[{0^ + }]}}{{\sin ({0^ + })}} = 0\)

بزرگترین توان

\(\begin{array}{l}1)\,\mathop {\lim }\limits_{x \to + \infty } f(x) = + 4\\\\2)\,\mathop {\lim }\limits_{x \to {{( - 3)}^ + }} f(x) = + \infty \end{array}\)

\(\begin{array}{l}g(x) = - {x^4} + m{x^3} - x + n\\\\g(2) = 5 \Rightarrow - 16 + 8m - 2 + n = 5 \Rightarrow 8m + n = + 23\\\\g( - 1) = 2 \Rightarrow - 1 - m + 1 + n = 2 \Rightarrow - m + n = 2\\\\ \Rightarrow \left\{ \begin{array}{l}8m = + 23\\\\m = - 2\end{array} \right.9m = 21 \Rightarrow m = \frac{{21}}{9} = \frac{7}{3}\\\\ \Rightarrow \frac{7}{3} - n = - 2 \Rightarrow n = \frac{7}{3} + 2 = \frac{{13}}{3}\end{array}\)

\(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {3^ + }} \frac{1}{{x - 3}} = + \infty \\\\\mathop {\lim }\limits_{x \to {3^ - }} \frac{1}{{x - 3}} = - \infty \end{array} \right. \Rightarrow \otimes \)

\(\mathop {\lim }\limits_{x \to - \infty } \frac{{7{x^8} + 2{x^3} - 2}}{{ - 14{x^5} + 3x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{7{x^8}}}{{ - 14{x^5}}} = - \frac{1}{2}{( - \infty )^3} = + \infty \)

\(\mathop {\lim }\limits_{x \to {{(\frac{\pi }{2})}^ + }} \tan x = - \infty \)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \frac{{ - 5x + 3}}{{3x - \sqrt {{x^2} - 3} }} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 5x + 3}}{{3x - |x|}}\\\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 5x + 3}}{{3x - ( - x)}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 5x}}{{4x}} = - \frac{5}{4}\end{array}\)