جواب کاردرکلاس صفحه 68 درس 3 حسابان دوازدهم (حدهای نامتناهی_حد در بی نهایت)

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الف

: مجانب قائم

\(\begin{array}{l}{x^2} - 1\, = 0 \Rightarrow {x^2} = 1 \Rightarrow x = \pm 1\\\\\mathop {\lim }\limits_{x \to 1} \frac{{x + 1}}{{{x^2} - 1}} = \frac{2}{0} = \not \exists \\\\\mathop {\lim }\limits_{x \to - 1} \frac{{x + 1}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to - 1} \frac{{x + 1}}{{(x - 1)(x + 1)}} = \mathop {\lim }\limits_{x \to - 1} \frac{1}{{x - 1}} = - \frac{1}{2}\\\\ \Rightarrow x = 1\end{array}\)

: مجانب افقی

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

ب

: مجانب قائم

مجانب قائم ندارد

: مجانب افقی

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

مجانب افقی ندارد

پ

: مجانب قائم

\(\begin{array}{l}x + 1\, = 0 \Rightarrow x = - 1 \Rightarrow x = \pm 1\\\\\mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} + 1}}{{x + 1}} = \frac{2}{0} = \not \exists \\\\ \Rightarrow x = - 1\end{array}\)

: مجانب افقی

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

مجانب افقی ندارد