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Câu hỏi: Tìm các giới hạn sau :
Lời giải chi tiết:
\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {{{x^3} - 5} \over {{x^2} + 1}} = \mathop {\lim }\limits_{x \to + \infty } {x}{{{x^2}\left({1 - {5 \over {{x^3}}}} \right)} \over {{x^2}\left({1 + {1 \over {{x^2}}}} \right)}} \cr
& = \mathop {\lim }\limits_{x \to + \infty } x.{{1 - {5 \over {{x^3}}}} \over {1 + {1 \over {{x^2}}}}} = + \infty \cr
& \text{vì} \mathop {\lim }\limits_{x \to + \infty } x = + \infty \text{và} \mathop {\lim }\limits_{x \to + \infty } {{1 - {5 \over {{x^3}}}} \over {1 + {1 \over {{x^2}}}}} = 1 > 0 \cr} \)
Cách khác:
a) Ta có:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3} - 5}}{{{x^2} + 1}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}\left( {1 - \frac{5}{{{x^3}}}} \right)}}{{{x^3}\left( {\frac{1}{x} + \frac{1}{{{x^2}}}} \right)}}\\
= \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \frac{5}{{{x^3}}}}}{{\frac{1}{x} + \frac{1}{{{x^2}}}}} = + \infty
\end{array}$
Vì $\mathop {\lim }\limits_{x \to \infty } \left( {1 - \frac{5}{{{x^3}}}} \right) = 1 - 0 = 1;\mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x} + \frac{1}{{{x^2}}}} \right) = 0$
và $\frac{1}{x} + \frac{1}{{{x^2}}} > 0;\forall x > 0$
Lời giải chi tiết:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^4} - x} }}{{1 - 2x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^4}\left({1 - \frac{1}{{{x^3}}}} \right)} }}{{1 - 2x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2}\sqrt {1 - \frac{1}{{{x^3}}}} }}{{x\left({\frac{1}{x} - 2} \right)}} = \mathop {\lim }\limits_{x \to - \infty } \left[ {x.\frac{{\sqrt {1 - \frac{1}{{{x^3}}}} }}{{\frac{1}{x} - 2}}} \right]
\end{array}\)
Ta có
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } x = - \infty \\
\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {1 - \frac{1}{{{x^3}}}} }}{{\frac{1}{x} - 2}} = \frac{1}{{ - 2}} < 0
\end{array}\)
Do đó \(\mathop {\lim }\limits_{x \to - \infty } \left( {x.\frac{{\sqrt {1 - \frac{1}{{{x^3}}}} }}{{\frac{1}{x} - 2}}} \right) = + \infty \)
Vậy \(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^4} - x} } \over {1 - 2x}}= + \infty \)
Cách khác:
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 \(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^4} - x} } \over {1 - 2x}} = + \infty \)
Câu a
\(\mathop {\lim }\limits_{x \to + \infty } {{{x^3} - 5} \over {{x^2} + 1}}\)Lời giải chi tiết:
\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {{{x^3} - 5} \over {{x^2} + 1}} = \mathop {\lim }\limits_{x \to + \infty } {x}{{{x^2}\left({1 - {5 \over {{x^3}}}} \right)} \over {{x^2}\left({1 + {1 \over {{x^2}}}} \right)}} \cr
& = \mathop {\lim }\limits_{x \to + \infty } x.{{1 - {5 \over {{x^3}}}} \over {1 + {1 \over {{x^2}}}}} = + \infty \cr
& \text{vì} \mathop {\lim }\limits_{x \to + \infty } x = + \infty \text{và} \mathop {\lim }\limits_{x \to + \infty } {{1 - {5 \over {{x^3}}}} \over {1 + {1 \over {{x^2}}}}} = 1 > 0 \cr} \)
Cách khác:
a) Ta có:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3} - 5}}{{{x^2} + 1}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}\left( {1 - \frac{5}{{{x^3}}}} \right)}}{{{x^3}\left( {\frac{1}{x} + \frac{1}{{{x^2}}}} \right)}}\\
= \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \frac{5}{{{x^3}}}}}{{\frac{1}{x} + \frac{1}{{{x^2}}}}} = + \infty
\end{array}$
Vì $\mathop {\lim }\limits_{x \to \infty } \left( {1 - \frac{5}{{{x^3}}}} \right) = 1 - 0 = 1;\mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x} + \frac{1}{{{x^2}}}} \right) = 0$
và $\frac{1}{x} + \frac{1}{{{x^2}}} > 0;\forall x > 0$
Câu b
\(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^4} - x} } \over {1 - 2x}}\)Lời giải chi tiết:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^4} - x} }}{{1 - 2x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^4}\left({1 - \frac{1}{{{x^3}}}} \right)} }}{{1 - 2x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2}\sqrt {1 - \frac{1}{{{x^3}}}} }}{{x\left({\frac{1}{x} - 2} \right)}} = \mathop {\lim }\limits_{x \to - \infty } \left[ {x.\frac{{\sqrt {1 - \frac{1}{{{x^3}}}} }}{{\frac{1}{x} - 2}}} \right]
\end{array}\)
Ta có
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } x = - \infty \\
\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {1 - \frac{1}{{{x^3}}}} }}{{\frac{1}{x} - 2}} = \frac{1}{{ - 2}} < 0
\end{array}\)
Do đó \(\mathop {\lim }\limits_{x \to - \infty } \left( {x.\frac{{\sqrt {1 - \frac{1}{{{x^3}}}} }}{{\frac{1}{x} - 2}}} \right) = + \infty \)
Vậy \(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^4} - x} } \over {1 - 2x}}= + \infty \)
Cách khác:
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 \(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^4} - x} } \over {1 - 2x}} = + \infty \)
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