Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to + \infty } {{{x^3} - 5} \over {{x^2} + 1}}\)
b. \(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^4} - x} } \over {1 - 2x}}\)
a.
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\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {{{x^3} - 5} \over {{x^2} + 1}} = \mathop {\lim }\limits_{x \to + \infty } {x}{{{x^2}\left( {1 - {5 \over {{x^3}}}} \right)} \over {{x^2}\left( {1 + {1 \over {{x^2}}}} \right)}} \cr
& = \mathop {\lim }\limits_{x \to + \infty } x.{{1 - {5 \over {{x^3}}}} \over {1 + {1 \over {{x^2}}}}} = + \infty \cr
& \text{vì}\,\mathop {\lim }\limits_{x \to + \infty } x = + \infty \,\text{và}\,\mathop {\lim }\limits_{x \to + \infty } {{1 - {5 \over {{x^3}}}} \over {1 + {1 \over {{x^2}}}}} = 1 > 0 \cr} \)
b.
Với mọi \(x < 0\), ta có \({{\sqrt {{x^4} - x} } \over {1 - 2x}} = {{{x^2}\sqrt {1 - {1 \over {{x^3}}}} } \over {1 - 2x}} = {{\sqrt {1 - {1 \over {{x^3}}}} } \over {{1 \over {{x^2}}} - {2 \over x}}}\)
Vì \(\mathop {\lim }\limits_{x \to - \infty } \sqrt {1 - {1 \over {{x^3}}}} = 1,\mathop {\lim }\limits_{x \to - \infty } \left( {{1 \over {{x^2}}} - {2 \over x}} \right) = 0\,\text{ và }\,{1 \over {{x^2}}} - {2 \over x} > 0\) với mọi \(x < 0\)
Nên \(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^4} - x} } \over {1 - 2x}} = + \infty \)