Bài 2.7 trang 164 Sách bài tập (SBT) Đại số và giải tích 11

Tính giới hạn của các hàm số sau khi \(x \to  + \infty \) và khi \(x \to  - \infty \)

a) \(f\left( x \right) = {{\sqrt {{x^2} - 3x} } \over {x + 2}}\) ;

b) \(f\left( x \right) = x + \sqrt {{x^2} - x + 1}\) ;

c) \(f\left( x \right) = \sqrt {{x^2} - x}  - \sqrt {{x^2} + 1} \) .

Giải:

a) Khi  \(x \to  + \infty \)

\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {{\sqrt {{x^2} - 3x} } \over {x + 2}} = \mathop {\lim }\limits_{x \to + \infty } {{\left| x \right|\sqrt {1 - {3 \over x}} } \over {x + 2}} \cr
& = \mathop {\lim }\limits_{x \to + \infty } {{x\sqrt {1 - {3 \over x}} } \over {x + 2}} = \mathop {\lim }\limits_{x \to + \infty } {{\sqrt {1 - {3 \over x}} } \over {1 + {2 \over x}}} = 1 \cr} \)

Khi \(x \to  - \infty \)

\(\eqalign{
& \mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^2} - 3x} } \over {x + 2}} = \mathop {\lim }\limits_{x \to - \infty } {{\left| x \right|\sqrt {1 - {3 \over x}} } \over {x + 2}} \cr
& = \mathop {\lim }\limits_{x \to - \infty } {{ - x\sqrt {1 - {3 \over x}} } \over {x + 2}} = \mathop {\lim }\limits_{x \to - \infty } {{ - \sqrt {1 - {3 \over x}} } \over {1 + {2 \over x}}} = - 1 \cr}\) ;

b)  Khi  \(x \to  + \infty \)

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

Khi \(x \to  - \infty \)

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

c) Khi  \(x \to  + \infty \)

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

Khi \(x \to  - \infty \)

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