Bài tập 4.35 trang 171 SBT Toán 10

Đặt \(g\left( x \right) = \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} - L\)

Ta có g(x) xác định trên (a;b)∖{x₀} và \(\mathop {\lim }\limits_{x \to {x_0}} g\left( x \right) = 0\)

Mặt khác, \(f\left( x \right) = f\left( {{x_0}} \right) + L.\left( {x - {x_0}} \right) + \left( {x - {x_0}} \right)g\left( x \right)\) nên 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = \mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( {{x_0}} \right) + L\left( {x - {x_0}} \right) + \left( {x - {x_0}} \right)g\left( x \right)} \right]\\
 = \mathop {\lim }\limits_{x \to {x_0}} f\left( {{x_0}} \right) + \mathop {\lim }\limits_{x \to {x_0}} L\left( {x - {x_0}} \right) + \mathop {\lim }\limits_{x \to {x_0}} \left( {x - {x_0}} \right).\mathop {\lim }\limits_{x \to {x_0}} g\left( x \right) = f\left( {{x_0}} \right)
\end{array}\)

Vậy hàm số liên tục tại x₀.

-- Mod Toán 11 HỌC247