Tìm \(m\) để hàm số \(f\left( x \right) = \left\{...

D

- Hướng dẫn giải

Phương pháp giải:

Xét tính liên tục của hàm số tại \(x = 0\).

Hàm số \(y = f\left( x \right)\)  liên tục tại điểm \(x = {x_0} \Leftrightarrow \mathop {\lim }\limits_{x \to x_0^ + } f\left( x \right) = \mathop {\lim }\limits_{x \to x_0^ - } f\left( x \right) = f\left( {{x_0}} \right).\)  

Giải chi tiết:

Hàm số liên tục trên \(\left( { - \infty ;0} \right);\left( {0; + \infty } \right)\).

Ta có: \(f\left( 0 \right) = m\,\,\,\)  

\(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{x - \tan x}}{{x - \sin x}} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x - \tan x}}{{{x^3}}}.\frac{{{x^3}}}{{x - \sin x}}} \right).\)

Đặt \(\left\{ \begin{array}{l}I = \mathop {\lim }\limits_{x \to 0} \frac{{x - \tan x}}{{{x^3}}}\\J = \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{{x^3}}}\end{array} \right.\)

Xét các giới hạn, đặt \(x = 3y\) ta có: \(\mathop {\lim }\limits_{x \to 0} \frac{x}{3} = 0 \Rightarrow x \to 0 \Rightarrow y \to 0.\)

\(\begin{array}{l}I = \mathop {\lim }\limits_{y \to 0} \frac{{3y - \tan 3y}}{{27{y^3}}} = \mathop {\lim }\limits_{y \to 0} \frac{{3y - \frac{{3\tan y - {{\tan }^3}y}}{{1 - 3{{\tan }^2}y}}}}{{27{y^3}}}\\ = \mathop {\lim }\limits_{y \to 0} \left[ {\frac{1}{{1 - 3{{\tan }^2}y}}.\frac{{3y\left( {1 - 3{{\tan }^2}y} \right) - 3\tan y + {{\tan }^3}y}}{{27{y^3}}}} \right]\\ = \mathop {\lim }\limits_{y \to 0} \frac{1}{{1 - 3{{\tan }^2}y}}.\mathop {\lim }\limits_{y \to 0} \left( {\frac{{3y - 3\tan y}}{{27{y^3}}} + \frac{{{{\tan }^3}y}}{{27{y^3}}} - \frac{{9y{{\tan }^2}y}}{{27{y^3}}}} \right)\\ = \mathop {\lim }\limits_{y \to 0} \left( {\frac{3}{{27}}.\frac{{y - \tan y}}{{{y^3}}} + \frac{1}{{27}}.\frac{{{{\tan }^3}y}}{{{y^3}}} - \frac{9}{{27}}.\frac{{y{{\tan }^2}y}}{{{y^3}}}} \right)\\ = \mathop {\lim }\limits_{y \to 0} \frac{3}{{27}}.\frac{{y - \tan y}}{{{y^3}}} + \frac{1}{{27}}\mathop {\lim }\limits_{y \to 0} \frac{{{{\sin }^3}y}}{{{y^3}.{{\cos }^3}y}} - \frac{9}{{27}}\mathop {\lim }\limits_{y \to 0} \frac{{y.{{\sin }^2}y}}{{{y^3}{{\cos }^2}y}}\\ = \frac{{3I}}{{27}} + \frac{1}{{27}} - \frac{1}{3}\\ \Leftrightarrow \frac{8}{9}I =  - \frac{8}{{27}} \Leftrightarrow I =  - \frac{1}{3}.\end{array}\)

\(\begin{array}{l}J = \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{{x^3}}} = \mathop {\lim }\limits_{y \to 0} \frac{{3y - \sin 3y}}{{27{y^3}}}\\ = \mathop {\lim }\limits_{y \to 0} \frac{{3y - \left( { - 4{{\sin }^3}y + 3\sin y} \right)}}{{27{y^3}}}\\ = \mathop {\lim }\limits_{y \to 0} \left( {\frac{4}{{27}}.\frac{{{{\sin }^3}y}}{{{y^3}}} + \frac{3}{{27}}.\frac{{y - \sin y}}{{{y^3}}}} \right)\\ = \frac{4}{{27}} + \frac{{3J}}{{27}}\\ \Leftrightarrow \frac{8}{9}J = \frac{4}{{27}} \Leftrightarrow J = \frac{1}{6}.\end{array}\)

\( \Rightarrow \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{x - \tan x}}{{x - \sin x}} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x - \tan x}}{{{x^3}}}.\frac{{{x^3}}}{{x - \sin x}}} \right) = \frac{I}{J} =  - \frac{1}{3}:\frac{1}{6} =  - 2.\)

\( \Rightarrow \) Hàm số liên tục trên \(\mathbb{R} \Leftrightarrow m =  - 2.\)

Chọn D.