Tìm các giới hạn sau:
LG a
limx→0e2−e3x+2x
Phương pháp giải:
Sử dụng giới hạn \mathop {\lim }\limits_{u \to 0} \frac{{{e^u} - 1}}{u} = 1
Lời giải chi tiết:
\mathop {\lim }\limits_{x \to 0} {{{e^2} - {e^{3x + 2}}} \over x} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^2} - {e^{3x}}.{e^2}}}{x}
= \mathop {\lim }\limits_{x \to 0} {{{-e^2}\left( {e^{3x}-1} \right)} \over x}= - {e^2}.\mathop {\lim }\limits_{x \to 0} \frac{{3\left( {{e^{3x}} - 1} \right)}}{{3x}}
= - 3{e^2}.\mathop {\lim }\limits_{x \to 0} {{{e^{3x}} - 1} \over {3x}} = - 3{e^2}.1=- 3{e^2} .
LG b
\mathop {\lim }\limits_{x \to 0} {{{e^{2x}} - {e^{5x}}} \over x}
Lời giải chi tiết:
\mathop {\lim }\limits_{x \to 0} {{{e^{2x}} - {e^{5x}}} \over x} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( {{e^{2x}} - 1} \right) - \left( {{e^{5x}} - 1} \right)}}{x}} \right)
\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} - 1}}{x} - \mathop {\lim }\limits_{x \to 0} \frac{{{e^{5x}} - 1}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {{e^{2x}} - 1} \right)}}{{2x}} - \mathop {\lim }\limits_{x \to 0} \frac{{5\left( {{e^{5x}} - 1} \right)}}{{5x}}\\ = 2\mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} - 1}}{{2x}} - 5\mathop {\lim }\limits_{x \to 0} \frac{{{e^{5x}} - 1}}{{5x}}\\ = 2.1 - 5.1\\ = - 3 \end{array}
