Tính$\mathop {\lim }\limits_{x \to  - \infty } (x - 1)\sqrt {\dfrac{{{x^2}}}{{2{x^4} + {x^2} + 1}}} $ bằng?

Giải thích thêm:

Cách khác: Đưa x ra ngoài căn bậc hai

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  - \infty } \left( {x - 1} \right)\sqrt {\dfrac{{{x^2}}}{{2{{\rm{x}}^4} + {x^2} + 1}}} \\ = \mathop {\lim }\limits_{x \to -\infty } \left( {x - 1} \right)\sqrt {\dfrac{1}{{{x^2}\left( {2 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^4}}}} \right)}}} \\ = \mathop {\lim }\limits_{x \to  - \infty } \left( {x - 1} \right)\dfrac{1}{{\left| x \right|}}\sqrt {\dfrac{1}{{2 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^4}}}}}} \\ = \mathop {\lim }\limits_{x \to  - \infty } \left( {x - 1} \right).\dfrac{1}{{ - x}}\sqrt {\dfrac{1}{{2 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^4}}}}}} \left( {D{\rm{o }}x < 0} \right)\\ = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{x - 1}}{{ - x}}\sqrt {\dfrac{1}{{2 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^4}}}}}} \\ = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{1 - \dfrac{1}{x}}}{{ - 1}}.\sqrt {\dfrac{1}{{2 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^4}}}}}} \\ =  - 1.\dfrac{1}{{\sqrt 2 }} =  - \dfrac{{\sqrt 2 }}{2}\end{array}\)

Vì \(\mathop {\lim }\limits_{x \to  - \infty } \sqrt {\dfrac{1}{{2 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^4}}}}}}  = \dfrac{1}{{\sqrt 2 }};\)\(\mathop {\lim }\limits_{x \to -\infty } \dfrac{{ - 1 + \dfrac{1}{x}}}{1} =  - 1\)