- LG a
- LG b
Tìm các giới hạn sau:
LG a
\[\mathop {\lim }\limits_{x \to 0} {{{e^2} - {e^{3x + 2}}} \over x}\]
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:
\[\mathop {\lim }\limits_{x \to 0} {{{e^{2x}} - {e^{5x}}} \over x} = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\left[ {{e^{2x}} - 1} \right] - \left[ {{e^{5x}} - 1} \right]}}{x}} \right]\]
\[\begin{array}{l}
= \mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} - 1}}{x} - \mathop {\lim }\limits_{x \to 0} \frac{{{e^{5x}} - 1}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{2\left[ {{e^{2x}} - 1} \right]}}{{2x}} - \mathop {\lim }\limits_{x \to 0} \frac{{5\left[ {{e^{5x}} - 1} \right]}}{{5x}}\\
= 2\mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} - 1}}{{2x}} - 5\mathop {\lim }\limits_{x \to 0} \frac{{{e^{5x}} - 1}}{{5x}}\\
= 2.1 - 5.1\\
= - 3
\end{array}\]