- LG a
- LG b
Tìm các giới hạn sau :
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 \[\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^4} - x} } \over {1 - 2x}} = + \infty \]