Phương pháp giải:

Nhân liên hợp.

Lời giải chi tiết:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  + \infty } \left( {x - \sqrt {{x^2} - 2x + 3} } \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{{x^2} - \left( {{x^2} - 2x + 3} \right)}}{{x + \sqrt {{x^2} - 2x + 3} }}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2x - 3}}{{x + \sqrt {{x^2} - 2x + 3} }}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{x\left( {2 - \dfrac{3}{x}} \right)}}{{x\left( {1 + \sqrt {1 - 2.\dfrac{1}{x} + \dfrac{1}{{{x^2}}}} } \right)}}\\ = \dfrac{2}{{1 + 1}} = 1\end{array}\)

Phương pháp giải:

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

Lời giải chi tiết:

\(\begin{array}{l}f\left( 2 \right) = 2m - 1\\\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {3x - 5}  - 1}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{3\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {\sqrt {3x - 5}  + 1} \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{3}{{\left( {\sqrt {3x - 5}  + 1} \right)}} = \dfrac{3}{2}\\ \Rightarrow 2m - 1 = \dfrac{3}{2} \Leftrightarrow m = \dfrac{5}{4}\end{array}\)