Đáp án:
\[\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x + \sqrt {x - 1} - 1}}{{\sqrt {{x^2} - 1} }} = \frac{1}{{\sqrt 2 }}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x + \sqrt {x - 1} - 1}}{{\sqrt {{x^2} - 1} }}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt x - 1} \right) + \sqrt {x - 1} }}{{\sqrt {{x^2} - 1} }}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{x - 1}}{{\sqrt x + 1}} + \sqrt {x - 1} }}{{\sqrt {x - 1} .\sqrt {x + 1} }}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{\sqrt {x - 1} }}{{\sqrt x + 1}} + 1}}{{\sqrt {x + 1} }}\\
= \frac{{\frac{{\sqrt {1 - 1} }}{{\sqrt 1 + 1}} + 1}}{{\sqrt {1 + 1} }}\\
= \frac{1}{{\sqrt 2 }}
\end{array}\)
