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 \(\mathop {\lim }\limits_{x \to {x_o}} \frac{{f(x)}}{{g(x)}} = + \infty \)
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\(\mathop {\lim }\limits_{x \to {x_o}} f(x) = L > 0\) 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) \(\mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sin x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{f(x)}}{x} = - \infty .\)