Tính \[\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x + 3} - x} \right)\]bằng?
A.−1.
B.0.
C.\(\frac{1}{2}\)
D.1
Bước 1:
\[\begin{array}{*{20}{l}}{\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x + 3} - x} \right)}\\{ = \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {\sqrt {{x^2} + x + 3} - x} \right)\left( {\sqrt {{x^2} + x + 3} + x} \right)}}{{\left( {\sqrt {{x^2} + x + 3} + x} \right)}}}\\{ = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} + x + 3 - {x^2}}}{{\sqrt {{x^2} + x + 3} + x}}}\\{ = \mathop {\lim }\limits_{x \to + \infty } \frac{{x + 3}}{{\sqrt {{x^2} + x + 3} + x}}}\end{array}\]
Bước 2:
\[ = \mathop {\lim }\limits_{x \to + \infty } \frac{{1 + \frac{3}{x}}}{{\sqrt {1 + \frac{1}{x} + \frac{3}{{{x^2}}}} + 1}}\]
Bước 3:
\[ = \frac{{1 + 0}}{{\sqrt {1 + 0 + 0} + 1}} = \frac{1}{2}\]
Đáp án cần chọn là: C
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