lim x → − ∞ √ x^2 − 9 bằng A. 0 .

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\(\mathop {\lim }\limits_{x \to - \infty } {\kern 1pt} \sqrt {{x^2} - 9} \) bằng

\(0\).

\( + \infty \).

\( - \infty \).

\(3\).

Giải thích

Đáp án đúng là: B

Ta có \(\mathop {\lim }\limits_{x \to - \infty } {\kern 1pt} \sqrt {{x^2} - 9} = \;\mathop {\lim }\limits_{x \to - \infty } {\kern 1pt} \left| x \right|\sqrt {1 - \frac{9}{{{x^2}}}} = + \infty \)

Vì \(\mathop {\lim }\limits_{x \to - \infty } {\kern 1pt} \left| x \right| = \mathop {\lim }\limits_{x \to - \infty } {\kern 1pt} \left( { - x} \right) = + \infty \;\;\)và \(\mathop {\lim }\limits_{x \to - \infty } {\kern 1pt} \sqrt {1 - \frac{9}{{{x^2}}}} = 1 > 0\).