Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\frac{{\sqrt {{x^2} - 4x + 4} }}{{x - 2}}\;\;khi\;x < 2\\m{x^2} - 3\;\;\;\;khi\;x \ge 2\end{array} \right.\)(m là tham số).

a) Khi m = 1 thì \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 1\).

b) Hàm số f(x) liên tục tại x = 2 khi m = 1.

c) f(2) = 4m – 3.

d) \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 1\).

 a) Khi m = 1 thì \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {{x^2} - 3} \right) = 4 - 3 = 1\).

b) Có f(2) = 4m – 3.

\(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {m{x^2} - 3} \right) = 4m - 3\);

\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\sqrt {{x^2} - 4x + 4} }}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\sqrt {{{\left( {x - 2} \right)}^2}} }}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{2 - x}}{{x - 2}} =  - 1\).

Để hàm số f(x) liên tục tại x = 2 thì \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = f\left( 2 \right)\)

Û 4m – 3 = −1 \( \Leftrightarrow m = \frac{1}{2}\).

c) f(2) = 4m – 3.

d) \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {m{x^2} - 3} \right) = 4m - 3\).

Đáp án: a) Đúng;   b) Sai;   c) Đúng;   d) Sai.