Câu 44 trang 167 SGK Đại số và Giải tích 11 Nâng cao

Tìm các giới hạn sau :

LG a

 \(\mathop {\lim }\limits_{x \to - \infty } x\sqrt {{{2{x^3} + x} \over {{x^5} - {x^2} + 3}}} \)

Giải chi tiết:

Với \(x < 0\), ta có :

\(\eqalign{
& x\sqrt {{{2{x^3} + x} \over {{x^5} - {x^2} + 3}}} = - \left| x \right|\sqrt {{{2{x^3} + x} \over {{x^5} - {x^2} + 3}}} \cr 
& = - \sqrt {{{{x^2}\left( {2{x^3} + x} \right)} \over {{x^5} - {x^2} + 3}}} = - \sqrt {{{2 + {1 \over {{x^2}}}} \over {1 - {1 \over {{x^3}}} + {1 \over {{x^5}}}}}} \cr} \)

Do đó :  \(\mathop {\lim }\limits_{x \to - \infty } x\sqrt {{{2{x^3} + x} \over {{x^5} - {x^2} + 3}}} = - \sqrt 2 \)

LG b

 \(\mathop {\lim }\limits_{x \to - \infty } {{\left| x \right| + \sqrt {{x^2} + x} } \over {x + 10}}\)

Giải chi tiết:

\(\eqalign{
& \mathop {\lim }\limits_{x \to - \infty } {{\left| x \right|+\sqrt {{x^2} + x} } \over {x + 10}} = \mathop {\lim }\limits_{x \to - \infty } {{\left| x \right| + \left| x \right|\sqrt {1 + {1 \over x}} } \over {x + 10}} \cr 
& = \mathop {\lim }\limits_{x \to - \infty } {{ - x - x\sqrt {1 + {1 \over x}} } \over {x + 10}} = \mathop {\lim }\limits_{x \to - \infty } {{ - 1 - \sqrt {1 + {1 \over x}} } \over {1 + {{10} \over x}}} \cr &= - 2 \cr} \)

LG c

\(\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {2{x^4} + {x^2} - 1} } \over {1 - 2x}}\)

Giải chi tiết:

\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {{\sqrt {2{x^4} + {x^2} - 1} } \over {1 - 2x}} = \mathop {\lim }\limits_{x \to + \infty } {{{x^2}\sqrt {2 + {1 \over {{x^2}}} - {1 \over {{x^4}}}} } \over {x\left( {{1 \over x} - 2} \right)}} \cr 
& = \mathop {\lim }\limits_{x \to + \infty } x{{\sqrt {2 + {1 \over {{x^2}}} - {1 \over {{x^4}}}} } \over {{1 \over x} - 2}} = - \infty \cr 
& \text{vì}\,\mathop {\lim }\limits_{x \to + \infty } x = + \infty \,\text{và}\,\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {2 + {1 \over {{x^2}}} - {1 \over {{x^4}}}} } \over {{1 \over x} - 2}} = - {{\sqrt 2 } \over 2} < 0 \cr} \)

LG d

 \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {2{x^2} + 1} + x} \right)\)

Giải chi tiết:

\(\eqalign{
& \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {2{x^2} + 1} + x} \right) = \mathop {\lim }\limits_{x \to - \infty } {{2{x^2} + x - {x^2}} \over {\sqrt {2{x^2} + x} - x}} \cr 
& = \mathop {\lim }\limits_{x \to - \infty } {{x\left( {x + 1} \right)} \over { - x\left( {\sqrt {2 + {1 \over x}} + 1} \right)}} \cr &= \mathop {\lim }\limits_{x \to - \infty } - {{x + 1} \over {\sqrt {2 + {1 \over x} + 1} }} = + \infty \cr 
& \text{vì }\,\mathop {\lim }\limits_{x \to - \infty } \left( { - x - 1} \right) = + \infty \cr} \)