Hoạt động 3 Bài 2 (trang 44) Toán 11: Cho biết mathop lim limits_x -> 0 sin x/x = 1. Dùng định nghĩa tính đạo hàm của hàm số y = sin x

Câu hỏi/Đề bài:

Cho biết \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\). Dùng định nghĩa tính đạo hàm của hàm số \(y = \sin x\).

Hướng dẫn:

Tính giới hạn \(f’\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) – f\left( {{x_0}} \right)}}{{x – {x_0}}}\).

Lời giải:

Với bất kì \({x_0} \in \mathbb{R}\), ta có:

\(f’\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) – f\left( {{x_0}} \right)}}{{x – {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sin x – \sin {x_0}}}{{x – {x_0}}}\)

Đặt \(x = {x_0} + \Delta x\). Ta có:

\(\begin{array}{l}f’\left( {{x_0}} \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \left( {{x_0} + \Delta x} \right) – \sin {x_0}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\cos \Delta x + \cos {x_0}\sin \Delta x – \sin {x_0}}}{{\Delta x}}\\ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\cos \Delta x – \sin {x_0}}}{{\Delta x}} + \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos {x_0}\sin \Delta x}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\left( {\cos \Delta x – 1} \right)}}{{\Delta x}} + \mathop {\lim }\limits_{\Delta x \to 0} \cos {x_0}.\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \Delta x}}{{\Delta x}}\end{array}\)

Lại có:

\(\begin{array}{l}\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\left( {\cos \Delta x – 1} \right)}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\left( {\cos \Delta x – 1} \right)\left( {\cos \Delta x + 1} \right)}}{{\Delta x\left( {\cos \Delta x + 1} \right)}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\left( {{{\cos }^2}\Delta x – 1} \right)}}{{\Delta x\left( {\cos \Delta x + 1} \right)}}\\ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\left( { – {{\sin }^2}\Delta x} \right)}}{{\Delta x\left( {\cos \Delta x + 1} \right)}} = – \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \Delta x}}{{\Delta x}}.\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}.\sin \Delta x}}{{\left( {\cos \Delta x + 1} \right)}} = – 1.\frac{{\sin {x_0}.\sin 0}}{{\cos 0 + 1}} = 0\\\mathop {\lim }\limits_{\Delta x \to 0} \cos {x_0}.\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \Delta x}}{{\Delta x}} = \cos {x_0}.1 = \cos {x_0}\end{array}\)

Vậy \(f’\left( {{x_0}} \right) = \cos {x_0}\)

Vậy \(f’\left( x \right) = \cos x\) trên \(\mathbb{R}\).