Bài 48 trang 112 SGK Đại số và Giải tích 12 Nâng cao

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:

\(\mathop {\lim }\limits_{x \to 0} {{{e^{2x}} – {e^{5x}}} \over x}  = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( {{e^{2x}} – 1} \right) – \left( {{e^{5x}} – 1} \right)}}{x}} \right)\)

\(\begin{array}{l}
= \mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} – 1}}{x} – \mathop {\lim }\limits_{x \to 0} \frac{{{e^{5x}} – 1}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {{e^{2x}} – 1} \right)}}{{2x}} – \mathop {\lim }\limits_{x \to 0} \frac{{5\left( {{e^{5x}} – 1} \right)}}{{5x}}\\
= 2\mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} – 1}}{{2x}} – 5\mathop {\lim }\limits_{x \to 0} \frac{{{e^{5x}} – 1}}{{5x}}\\
= 2.1 – 5.1\\
= – 3
\end{array}\)

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