جواب تمرین صفحه 57 درس 3 ریاضی دوازدهم تجربی (حد بی نهایت و حد در بی نهایت)

الف  بخش پذیر است: 

\(x + 1 = 0 \Rightarrow x = - 1 \Rightarrow f( - 1) = 2{( - 1)^3} + {( - 1)^2} + 1 = 0\)

ب

شکل

\( \Rightarrow f(x) = 2{x^3} + {x^2} + 1 = (x + 1)(\;2{x^2} - x + 1)\)

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

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

ب \(\mathop {\lim }\limits_{x \to 5} \;\frac{{{x^3} - 4{x^2} - 4x - 5}}{{{x^2} - 25}}\)

\( = \mathop {\lim }\limits_{x \to 5} \frac{{(x - 5)({x^2} + x + 1)}}{{(x - 5)(x + 5)}} = \mathop {\lim }\limits_{x \to 5} \frac{{({x^2} + x + 1)}}{{(x + 5)}} = \frac{{31}}{{10}}\)

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

\( = \mathop {\lim }\limits_{x \to \; - \;4} \frac{{(x + 4)(x - 1)}}{{(x + 4)({x^2} + 1)}} = \mathop {\lim }\limits_{x \to \; - \;4} \frac{{(x - 1)}}{{({x^2} + 1)}} = \frac{{ - \;5}}{{17}}\)

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

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \;\frac{{x - \sqrt {2x - 1} }}{{{x^2} - x}} \times \frac{{x + \sqrt {2x - 1} }}{{x + \sqrt {2x - 1} }} = \mathop {\lim }\limits_{x \to \;1} \frac{{{x^2} - 2x + 1}}{{x(x - 1)(x + \sqrt {2x - 1} )}}\\\\ = \mathop {\lim }\limits_{x \to \;1} \frac{{{{(x - 1)}^2}}}{{x(x - 1)(x + \sqrt {2x - 1} )}}{\kern 1pt} {\kern 1pt} = \mathop {\lim }\limits_{x \to \;1} \frac{{(x - 1)}}{{x(x + \sqrt {2x - 1} )}} = \frac{0}{2} = 0\end{array}\)

ب \(\mathop {\lim }\limits_{x \to 3} \;\frac{{{x^2} - 9}}{{2 - \sqrt {x + 1} }}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \;\frac{{{x^2} - 9}}{{2 - \sqrt {x + 1} }} \times \frac{{2 + \sqrt {x + 1} }}{{2 + \sqrt {x + 1} }} = \mathop {\lim }\limits_{x \to \;3} \frac{{(x - 3)(x + 3)(2 + \sqrt {x + 1} )}}{{ - (x - 3)}}\\\\ = \mathop {\lim }\limits_{x \to \;3} \frac{{(x + 3)(2 + \sqrt {x + 1} )}}{{ - 1}} = \frac{{6 \times 4}}{{ - 1}} = - 24\end{array}\)

پ \(\mathop {\lim }\limits_{x \to - 8} \;\frac{{2x + 16}}{{\sqrt[3]{x} + 2}}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to - 8} \;\frac{{2x + 16}}{{\sqrt[3]{x} + 2}} \times \frac{{(\sqrt[3]{{{x^2}}} - 2\sqrt[3]{x} + 4)}}{{(\sqrt[3]{{{x^2}}} - 2\sqrt[3]{x} + 4)}} = \mathop {\lim }\limits_{x \to \; - \;8} \frac{{2(x + 8)(\sqrt[3]{{{x^2}}} - 2\sqrt[3]{x} + 4)}}{{(x + 8)}}\\\\ = \mathop {\lim }\limits_{x \to \; - \;8} \frac{{2(\sqrt[3]{{{x^2}}} - 2\sqrt[3]{x} + 4)}}{1} = 2 \times 4 = 8\end{array}\)

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

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

ب \(\mathop {\lim }\limits_{x \to 0} \;\frac{{ - 1}}{{\left| x \right|}}\)

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

پ \(\mathop {\lim }\limits_{x \to {1^ - }} \;\frac{1}{{x - 1}}\)

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

ت \(\mathop {\lim }\limits_{x \to - 6} \;\frac{9}{{{{\left( {x + 6} \right)}^2}}}\)

\(\mathop {\lim }\limits_{x \to - 6} \;\frac{9}{{{{\left( {x + 6} \right)}^2}}} = \frac{9}{{{0^ + }}} = + \;\infty \)

ث \(\mathop {\lim }\limits_{x \to 3} \frac{{ - 1}}{{{{\left( {x - 3} \right)}^4}}}\)

\(\mathop {\lim }\limits_{x \to 3} \frac{{ - 1}}{{{{\left( {x - 3} \right)}^4}}} = \frac{{ - \;1}}{{{0^ + }}} = - \;\infty \)

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

\(\mathop {\lim }\limits_{x \to \frac{{ - 1}}{2}} \;\frac{{4x + 1}}{{{{\left( {2x + 1} \right)}^2}}} = \frac{{ - 1}}{{{0^ + }}} = - \;\infty \)

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

\(\mathop {\lim }\limits_{x \to {3^ + }} \;\frac{{1 - 5x}}{{{x^2} - 9}} = \frac{{ - \;14}}{{{0^ + }}} = - \;\infty \)

ح \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} \;\frac{{ - 3x}}{{{x^2} - 4}}\)

\(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} \;\frac{{ - 3x}}{{{x^2} - 4}} = \frac{6}{{{0^ + }}} = + \;\infty \)

خ \(\mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ + }} \;\frac{1}{{\cos x}}\)

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

د \(\mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} \;\tan x\)

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

ذ \(\mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ + }} \;\tan x\)

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

ر \(\mathop {\lim }\limits_{x \to {3^ - }} \;\frac{{\left[ x \right] - 3}}{{x - 3}}\)

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

الف حد تابع f(x) وقتی x از مقادیر کوچکتر از 2 به 2 نزدیک تر می شود، از هر عدد مثبت دلخواهی بزرگتر است.

ب حد تابع f(x) وقتی x از مقادیر بزرگتر از 2 به 2 نزدیک تر می شود، از هر عدد منفی دلخواهی کوچکتر است.

پ

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