- LG a
- LG b
- LG c
- LG d
Tìm các giới hạn sau:
LG a
\[\mathop {\lim }\limits_{x \to 0} {{{e^{3x}} - 1} \over x}\]
Lời giải chi tiết:
\[\mathop {\lim }\limits_{x \to 0} {{{e^{3x}} - 1} \over x}\]
\[= 3.\mathop {\lim }\limits_{x \to 0} {{{e^{3x}} - 1} \over {3x}} = 3.1 = 3\]
LG b
\[\mathop {\lim }\limits_{x \to 0} {{{e^{2x}} - {e^{3x}}} \over {5x}}\]
Lời giải chi tiết:
\[\eqalign{ & \mathop {\lim }\limits_{x \to 0} {{{e^{2x}} - {e^{3x}}} \over {5x}} = \mathop {\lim }\limits_{x \to 0} \left[ {{{{e^{2x }-1}} \over {5x}} - {{{e^{3x }-1}} \over {5x}}} \right] \cr& = \mathop {\lim }\limits_{x \to 0} {{{e^{2x }-1}} \over {2x}}.{2 \over 5} - \mathop {\lim }\limits_{x \to 0} {{{e^{3x }-1}} \over {3x}}.{3 \over 5} \cr&= {2 \over 5} - {3 \over 5} = - {1 \over 5} \cr} \]
LG c
\[\mathop {\lim }\limits_{x \to 5} \left[ {{2^x} - {3^x}} \right]\]
Lời giải chi tiết:
\[\mathop {\lim }\limits_{x \to 5} \left[ {{2^x} - {3^x}} \right]\]
\[= {2^5} - {3^5} = - 211\]
LG d
\[\mathop {\lim }\limits_{x \to + \infty } \left[ {x{e^{{1 \over x}}} - x} \right]\]
Lời giải chi tiết:
\[\mathop {\lim }\limits_{x \to + \infty } \left[ {x{e^{{1 \over x}}} - x} \right] = \mathop {\lim }\limits_{x \to + \infty } {{{e^{{1 \over x} }-1}} \over {{1 \over x}}} = \mathop {\lim }\limits_{y \to 0^+ } {{{e^y} - 1} \over y} = 1\]