جواب تمرین صفحه 58 درس 3 حسابان دوازدهم (حدهای نامتناهی_حد در بی نهایت)

الف \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {3 + {x^2}} }}{{{x^2}}} = + \infty \Rightarrow \left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to 0} \sqrt {3 + {x^2}} = \sqrt 3 \\\\\mathop {\lim }\limits_{x \to 0} {x^2} = {0^ + }\end{array} \right.\quad \quad \)

: طبق قضیه 3

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

ب \(\mathop {\lim }\limits_{x \to 2} \frac{1}{{{{\left( {x - 2} \right)}^4}}} = + \infty \Rightarrow \left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to 2} (1) = 1\\\\\mathop {\lim }\limits_{x \to 2} {(x - 2)^4} = {0^ + }\end{array} \right.\quad \quad \)

: طبق قضیه 3

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

پ \(\mathop {\lim }\limits_{x \to - 2} \left| {\frac{{5 - x}}{{2 + x}}} \right| = + \infty \Rightarrow \left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to ( - 2)} \left| {5 - x} \right| = 7\\\\\mathop {\lim }\limits_{x \to ( - 2)} \left| {2 + x} \right| = {0^ + }\end{array} \right.\quad \)

: طبق قضیه 3

\(\mathop {\lim }\limits_{x \to 0} \left| {\frac{{5 - x}}{{2 + x}}} \right| = \frac{7}{{{0^ + }}} = + \;\infty \quad \)

الف \(\mathop {\lim }\limits_{x \to {2^ - }} \frac{{2x}}{{{x^2} - 4}} = \frac{{2(2)}}{{{{({2^ - })}^2} - 4}} = \frac{4}{{{0^ - }}} = - \;\infty \)

ب \(\mathop {\lim }\limits_{x \to {3^ - }} \frac{{{x^2} + 2x - 1}}{{{x^2} + x - 12}} = \frac{{{{(3)}^2} + 2(3) - 1}}{{{{({3^ - })}^2} + ({3^ - }) - 12}} = \frac{{14}}{{{0^ - }}} = - \;\infty \)

پ \(\mathop {\lim }\limits_{x \to {3^ + }} \frac{{x + 1}}{{9 - {x^2}}} = \frac{{3 + 1}}{{9 - {{({3^ + })}^2}}} = \frac{4}{{{0^ - }}} = - \;\infty \)

الف \(f\left( x \right) = \frac{{2x - 1}}{{3 - x}}\)

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

ب \(g\left( x \right) = \frac{{{x^2} + x}}{{{x^2} - x}}\)

\(\begin{array}{l}{x^2} - x = 0 \Rightarrow \left\{ \begin{array}{l}x = 0\\\\x = 1\end{array} \right.\\\\ \Rightarrow x = 1\\\\\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + x}}{{{x^2} - x}} = \frac{2}{0} = \infty \end{array}\)

x=0 مجانب نیست؛ زیرا تابع در این نقطه دارای حد می باشد:

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

شکل (الف) ؛ زیرا :

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