شاه کلید نکات
شاه کلید سوال
حد های زیر را محاسبه کنید.
الف \(\mathop {\lim }\limits_{x \to - {2^ + }} \frac{{1 - x}}{{x + 2}}\)
ب \(\mathop {\lim }\limits_{x \to {2^ - }} \frac{{\left[ x \right] - 2}}{{x - 2}}\)
پ \(\mathop {\lim }\limits_{x \to {2^ - }} \frac{{\left[ x \right] - 2}}{{x - 2}}\)
الف \(\mathop {\lim }\limits_{x \to - {2^ + }} \frac{{1 - x}}{{x + 2}} = \frac{{1 - ( - 2)}}{{( - {2^ + }) + 2}} = \frac{3}{{{0^ + }}} = + \;\infty \)
ب \(\mathop {\lim }\limits_{x \to {2^ - }} \frac{{\left[ x \right] - 2}}{{x - 2}} = \frac{{\left[ {{2^ - }} \right] - 2}}{{({2^ - }) - 2}} = \)
\(\frac{{1 - 2}}{{{0^ - }}} = \frac{{ - 1}}{{{0^ - }}} = + \;\infty \)
پ \(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{x^2} - 1}}{{{{(x - 1)}^2}}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{(x + 1)(x - 1)}}{{{{(x - 1)}^2}}} = \)
\(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{(x + 1)}}{{(x - 1)}} = \frac{{1 + 1}}{{{1^ + } - 1}} = \frac{2}{{{0^ + }}} = + \;\infty \)
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