Bài tập 37 trang 163 SGK Toán 11 NC

a) Ta có \(\mathop {\lim }\limits_{x \to 1} \frac{2}{{{{\left( {x - 1} \right)}^2}}} =  + \infty \) và \(\mathop {\lim }\limits_{x \to 1} \frac{{2x + 1}}{{2x - 3}} = \frac{3}{{ - 1}} =  - 3 < 0\)

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

b) Ta có \(\frac{5}{{\left( {x - 1} \right)\left( {{x^2} - 3x + 2} \right)}} = \frac{1}{{{{\left( {x - 1} \right)}^2}}}.\frac{5}{{x - 2}}\)

Vì \(\mathop {\lim }\limits_{x \to 1} \frac{1}{{{{\left( {x - 1} \right)}^2}}} =  + \infty ,\mathop {\lim }\limits_{x \to 1} \frac{5}{{x - 2}} =  - 5 < 0\)

Do đó \(\mathop {\lim }\limits_{x \to 1} \frac{5}{{\left( {x - 1} \right)\left( {{x^2} - 3x + 2} \right)}} =  - \infty \)

-- Mod Toán 11 HỌC247