Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{_{x \to 0}} \frac{{{e^2} - {e^{3x}} + 2}}{x}\)
b) \(\mathop {\lim }\limits_{_{x \to 0}} \frac{{{e^{2x}} - {e^{5x}}}}{x}\)
a)
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{{e^2} - {e^{3x + 2}}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^2}\left( {1 - {3^{3x}}} \right)}}{x}\\
= - 3{e^2}.\mathop {\lim }\limits_{x \to 0} \frac{{{e^{3x}} - 1}}{{3x}} = - 3{e^2}
\end{array}\)
b)
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} - {e^{5x}}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{e^{2x}} - 1}}{x} - \frac{{{e^{5x}} - 1}}{x}} \right)\\
= 2 - 5 = - 3
\end{array}\)
-- Mod Toán 12
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