Câu 36 trang 163 SGK Đại số và Giải tích 11 Nâng cao

LG 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:

LG 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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