- LG a
- LG b
- LG c
- LG d
- LG e
- LG f
- LG g
- LG h
Tính các giới hạn sau
LG a
\[\mathop {\lim }\limits_{x \to - 3} {{x + 3} \over {{x^2} + 2x - 3}}\]
Lời giải chi tiết:
\[\mathop {\lim }\limits_{x \to - 3} {{x + 3} \over {{x^2} + 2x - 3}} = \mathop {\lim }\limits_{x \to - 3} {{x + 3} \over {\left[ {x - 1} \right]\left[ {x + 3} \right]}} = \mathop {\lim }\limits_{x \to - 3} {1 \over {x - 1}} = {{ - 1} \over 4}\]
LG b
\[\mathop {\lim }\limits_{x \to + \infty } {{x - 1} \over {{x^2} - 1}}\]
Lời giải chi tiết:
\[\mathop {\lim }\limits_{x \to + \infty } {{x - 1} \over {{x^2} - 1}} = \mathop {\lim }\limits_{x \to + \infty } {{{1 \over x} - {1 \over {{x^2}}}} \over {1 - {1 \over {{x^2}}}}} = 0\]
LG c
\[\mathop {\lim }\limits_{x \to 5} {{x - 5} \over {\sqrt x - \sqrt 5 }}\]
Lời giải chi tiết:
\[\mathop {\lim }\limits_{x \to 5} {{x - 5} \over {\sqrt x - \sqrt 5 }}\]
\[= \mathop {\lim }\limits_{x \to 5} {{\left[ {\sqrt x - \sqrt 5 } \right]\left[ {\sqrt x + \sqrt 5 } \right]} \over {\sqrt x - \sqrt 5 }}\]
\[= \mathop {\lim }\limits_{x \to 5} \left[ {\sqrt x + \sqrt 5 } \right] = 2\sqrt 5 \]
LG d
\[\mathop {\lim }\limits_{x \to + \infty } {{x - 5} \over {\sqrt x + \sqrt 5 }}\]
Lời giải chi tiết:
\[\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {{x - 5} \over {\sqrt x + \sqrt 5 }} \cr
& = \mathop {\lim }\limits_{x \to + \infty } {{1 - {5 \over x}} \over {{1 \over {\sqrt x }} + {{\sqrt 5 } \over x}}} = + \infty \cr} \]
[Vì \[{1 \over {\sqrt x }} + {{\sqrt 5 } \over x} > 0\]với mọi \[x > 0\]]
LG e
\[\mathop {\lim }\limits_{x \to 1} {{\sqrt x - 1} \over {\sqrt {x + 3} - 2}}\]
Lời giải chi tiết:
\[\eqalign{
& \mathop {\lim }\limits_{x \to 1} {{\sqrt x - 1} \over {\sqrt {x + 3} - 2}} \cr
& = \mathop {\lim }\limits_{x \to 1} {{\left[ {\sqrt x - 1} \right]\left[ {\sqrt {x + 3} + 2} \right]} \over {x + 3 - 4}} \cr
& = \mathop {\lim }\limits_{x \to 1} {{\left[ {\sqrt {x - 1} } \right]\left[ {\sqrt {x + 3} + 2} \right]} \over {x - 1}} \cr
& = \mathop {\lim }\limits_{x \to 1} {{\left[ {\sqrt x - 1} \right]\left[ {\sqrt {x + 3} + 2} \right]} \over {\left[ {\sqrt x - 1} \right]\left[ {\sqrt x + 1} \right]}} \cr
& = \mathop {\lim }\limits_{x \to 1} {{\sqrt {x + 3} + 2} \over {\sqrt x + 1}} = 2 \cr} \]
LG f
\[\mathop {\lim }\limits_{x \to + \infty } {{1 - 2x + 3{x^3}} \over {{x^3} - 9}}\]
Lời giải chi tiết:
\[\mathop {\lim }\limits_{x \to + \infty } {{1 - 2x + 3{x^3}} \over {{x^3} - 9}} = \mathop {\lim }\limits_{x \to + \infty } {{{1 \over {{x^3}}} - {2 \over {{x^2}}} + 3} \over {1 - {9 \over {{x^3}}}}} = 3\]
LG g
\[\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{1 \over {{x^2} + 1}} - 1} \right]\]
Lời giải chi tiết:
\[\eqalign{
& \mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{1 \over {{x^2} + 1}} - 1} \right] \cr
& = \mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}.\left[ {{{ - {x^2}} \over {{x^2} + 1}}} \right] \cr
& = \mathop {\lim }\limits_{x \to 0} {{ - 1} \over {{x^2} + 1}} = - 1 \cr} \]
LG h
\[\mathop {\lim }\limits_{x \to - \infty } {{\left[ {{x^2} - 1} \right]{{\left[ {1 - 2x} \right]}^5}} \over {{x^7} + x + 3}}\]
Lời giải chi tiết:
\[\eqalign{
& \mathop {\lim }\limits_{x \to - \infty } {{\left[ {{x^2} - 1} \right]{{\left[ {1 - 2x} \right]}^5}} \over {{x^7} + x + 3}} \cr
& = \mathop {\lim }\limits_{x \to - \infty } {{{x^2}\left[ {1 - {1 \over {{x^2}}}} \right].{x^5}{{\left[ {{1 \over x} - 2} \right]}^5}} \over {{x^7} + x + 3}} \cr
& = \mathop {\lim }\limits_{x \to - \infty } {{\left[ {1 - {1 \over {{x^2}}}} \right]{{\left[ {{1 \over x} - 2} \right]}^5}} \over {1 + {1 \over {{x^6}}} + {3 \over {{x^7}}}}} \cr
& = {\left[ { - 2} \right]^5} = - 32 \cr}\]