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