- LG a
- LG b
- LG c
- LG d
Tìm các giới hạn sau :
LG a
\[\mathop {\lim }\limits_{x \to - \infty } x\sqrt {{{2{x^3} + x} \over {{x^5} - {x^2} + 3}}} \]
Lời giải chi tiết:
Với \[x < 0\], ta có :
\[ x\sqrt {{{2{x^3} + x} \over {{x^5} - {x^2} + 3}}}\]
\[\eqalign{
&= x\sqrt {\frac{{{x^3}\left[ {2 + \frac{1}{{{x^2}}}} \right]}}{{{x^5}\left[ {1 - \frac{1}{{{x^3}}} + \frac{3}{{{x^5}}}} \right]}}} \cr &= x\sqrt {\frac{{2 + \frac{1}{{{x^2}}}}}{{{x^2}\left[ {1 - \frac{1}{{{x^3}}} + \frac{3}{{{x^5}}}} \right]}}} \cr & = x.\frac{1}{{\left| x \right|}}.\sqrt {\frac{{2 + \frac{1}{{{x^2}}}}}{{1 - \frac{1}{{{x^3}}} + \frac{3}{{{x^5}}}}}} \cr
& = x.\frac{1}{{ - x}}.\sqrt {\frac{{2 + \frac{1}{{{x^2}}}}}{{1 - \frac{1}{{{x^3}}} + \frac{3}{{{x^5}}}}}} \cr &= - \sqrt {{{2 + {1 \over {{x^2}}}} \over {1 - {1 \over {{x^3}}} + {1 \over {{x^5}}}}}} \cr} \]
Do đó : \[\mathop {\lim }\limits_{x \to - \infty } x\sqrt {{{2{x^3} + x} \over {{x^5} - {x^2} + 3}}} = - \sqrt 2 \]
LG b
\[\mathop {\lim }\limits_{x \to - \infty } {{\left| x \right| + \sqrt {{x^2} + x} } \over {x + 10}}\]
Phương pháp giải:
Đưa \[x^2\] ra ngoài dấu căn, chú ý dấu của x.
Lời giải chi tiết:
\[\eqalign{
& \mathop {\lim }\limits_{x \to - \infty } {{\left| x \right|+\sqrt {{x^2} + x} } \over {x + 10}}\cr &= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left| x \right| + \sqrt {{x^2}\left[ {1 + \frac{1}{x}} \right]} }}{{x + 10}}\cr &= \mathop {\lim }\limits_{x \to - \infty } {{\left| x \right| + \left| x \right|\sqrt {1 + {1 \over x}} } \over {x + 10}} \cr
& = \mathop {\lim }\limits_{x \to - \infty } {{ - x - x\sqrt {1 + {1 \over x}} } \over {x + 10}} \cr &= \mathop {\lim }\limits_{x \to - \infty } {{ - 1 - \sqrt {1 + {1 \over x}} } \over {1 + {{10} \over x}}} \cr &= \frac{{ - 1 - \sqrt {1 + 0} }}{{1 + 0}}= - 2 \cr} \]
LG c
\[\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {2{x^4} + {x^2} - 1} } \over {1 - 2x}}\]
Lời giải chi tiết:
\[\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {{\sqrt {2{x^4} + {x^2} - 1} } \over {1 - 2x}} \cr &= \mathop {\lim }\limits_{x \to + \infty } {{{x^2}\sqrt {2 + {1 \over {{x^2}}} - {1 \over {{x^4}}}} } \over {x\left[ {{1 \over x} - 2} \right]}} \cr
& = \mathop {\lim }\limits_{x \to + \infty } x{{\sqrt {2 + {1 \over {{x^2}}} - {1 \over {{x^4}}}} } \over {{1 \over x} - 2}} = - \infty \cr
& \text{vì}\,\mathop {\lim }\limits_{x \to + \infty } x = + \infty \cr &\text{và}\,\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {2 + {1 \over {{x^2}}} - {1 \over {{x^4}}}} } \over {{1 \over x} - 2}} = - {{\sqrt 2 } \over 2} < 0 \cr} \]
Cách khác:
LG d
\[\mathop {\lim }\limits_{x \to - \infty } \left[ {\sqrt {2{x^2} + 1} + x} \right]\]
Phương pháp giải:
Nhân và chia với biểu thức \[\left[ {\sqrt {2{x^2} + 1} + x} \right]\]
Lời giải chi tiết:
\[\eqalign{
& \mathop {\lim }\limits_{x \to - \infty } \left[ {\sqrt {2{x^2} + 1} + x} \right]\cr & = \mathop {\lim }\limits_{x \to - \infty } {{2{x^2} + 1 - {x^2}} \over {\sqrt {2{x^2} + 1} - x}} \cr} \]
\[ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^2} + 1}}{{\sqrt {{x^2}\left[ {2 + \dfrac{1}{x^2}} \right]} - x}}\]
\[ = \mathop {\lim }\limits_{x \to - \infty } x.\dfrac{{x + \dfrac{1}{x}}}{{\left| x \right|\sqrt {2 + \dfrac{1}{x^2}} - x}} \]
\[ = \mathop {\lim }\limits_{x \to - \infty } x.\dfrac{{x + \dfrac{1}{x}}}{{ - x\sqrt {2 + \dfrac{1}{x^2}} - x}} \]
\[ = \mathop {\lim }\limits_{x \to - \infty } x.\dfrac{{1 + \dfrac{1}{x^2}}}{{ - \sqrt {2 + \dfrac{1}{x^2}} - 1}} = + \infty \]
Vì\[\mathop {\lim }\limits_{x \to - \infty } x = - \infty \] và\[\mathop {\lim }\limits_{x \to - \infty } \dfrac{{1 + \dfrac{1}{x^2}}}{{ - \sqrt {2 + \dfrac{1}{x}} - 1}} = \dfrac{1}{{ - \sqrt 2 - 1}} < 0\]