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