Đáp án:
\(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 2x - 1} - x + 2} \right) = 3\)
Giải thích các bước giải:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 2x - 1} - x + 2} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2} + 2x - 1 - {{\left( {x - 2} \right)}^2}}}{{\sqrt {{x^2} + 2x - 1} + x - 2}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2} + 2x - 1 - {x^2} + 4x - 4}}{{\sqrt {{x^2} + 2x - 1} + x - 2}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{6x - 5}}{{\sqrt {{x^2} + 2x - 1} + x - 2}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{6 - \dfrac{5}{x}}}{{\sqrt {1 + \dfrac{2}{x} - \dfrac{1}{{{x^2}}}} + 1 - \dfrac{2}{x}}} = \dfrac{6}{{1 + 1}} = 3
\end{array}\)
