Bài 5.48 trang 90 SBT Toán 11 - Kết nối tri thức: Biết $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$. Hãy tính...

Biết $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$. Hãy tính:

a) $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{{x^3}}}$;

b) $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x}}{{{x^2}}}$

c) $\mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sin x}}{{{x^2}}}$.

Áp dụng lý thuyết:

$\mathop {\lim }\limits_{x \to {x_o}} f(x) = L > 0$ và $\mathop {\lim }\limits_{x \to {x_o}} g(x) = 0$ thì $\mathop {\lim }\limits_{x \to {x_o}} \frac{{f(x)}}{{g(x)}} = + \infty$

$\mathop {\lim }\limits_{x \to {x_o}} f(x) = L > 0$ và $\mathop {\lim }\limits_{x \to {x_o}^ + } g(x) = 0$ thì $\mathop {\lim }\limits_{x \to {x_o}} \frac{{f(x)}}{{g(x)}} = + \infty$

$\mathop {\lim }\limits_{x \to {x_o}} f(x) = L > 0$ và $\mathop {\lim }\limits_{x \to x_{_o}^ - } g(x) = 0$ thì $\mathop {\lim }\limits_{x \to {x_o}} \frac{{f(x)}}{{g(x)}} = - \infty$

Đặt $f(x) = \frac{{\sin x}}{x}$. Khi đó

a) $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \frac{{f(x)}}{{{x^2}}} = + \infty .$

b) $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(x)}}{x} = + \infty$.

c) $\mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sin x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{f(x)}}{x} = - \infty .$