Nội dung bài giảng
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to 0} {{{e^{3x}} - 1} \over x}\) b) \(\mathop {\lim }\limits_{x \to 0} {{{e^{2x}} - {e^{3x}}} \over {5x}}\)
c) \(\mathop {\lim }\limits_{x \to 5} \left( {{2^x} - {3^x}} \right)\) d) \(\mathop {\lim }\limits_{x \to + \infty } \left( {x{e^{{1 \over x}}} - x} \right)\)
Giải
a) \(\mathop {\lim }\limits_{x \to 0} {{{e^{3x}} - 1} \over x}\)
\( = 3.\mathop {\lim }\limits_{x \to 0} {{{e^{3x}} - 1} \over {3x}} = 3.1 = 3\)
b)
\(\eqalign{ & \mathop {\lim }\limits_{x \to 0} {{{e^{2x}} - {e^{3x}}} \over {5x}} = \mathop {\lim }\limits_{x \to 0} \left( {{{{e^{2x }-1}} \over {5x}} - {{{e^{3x }-1}} \over {5x}}} \right) \cr& = \mathop {\lim }\limits_{x \to 0} {{{e^{2x }-1}} \over {2x}}.{2 \over 5} - \mathop {\lim }\limits_{x \to 0} {{{e^{3x }-1}} \over {3x}}.{3 \over 5} \cr&= {2 \over 5} - {3 \over 5} = - {1 \over 5} \cr} \)
c) \(\mathop {\lim }\limits_{x \to 5} \left( {{2^x} - {3^x}} \right)\)
\( = {2^5} - {3^5} = - 211\)
d)
\(\mathop {\lim }\limits_{x \to + \infty } \left( {x{e^{{1 \over x}}} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } {{{e^{{1 \over x} }-1}} \over {{1 \over x}}} = \mathop {\lim }\limits_{y \to 0^+ } {{{e^y} - 1} \over y} = 1\)