Giải bài 48 trang 112 SGK giải tích nâng cao 12

Áp dụng công thức: \(\lim\limits_{x\to 0}\,\dfrac{{{e}^{x}}-1}{x}=1\)

a) Ta có

\(\lim\limits_{x\to 0}\,\dfrac{{{e}^{2}}-{{e}^{3x+2}}}{x}=\lim\limits_{x\to 0}\,\dfrac{{{e}^{2}}\left( 1-{{e}^{3x}} \right)}{\dfrac{1}{3}.3x}=3{{e}^{2}}\)

b) Ta có

\(\lim\limits_{x\to 0}\,\dfrac{{{e}^{2x}}-{{e}^{5x}}}{x}=\lim\limits_{x\to 0}\,\dfrac{{{e}^{2x}}-1+1-{{e}^{5x}}}{x} \\ =2\lim\limits_{x\to 0}\,\dfrac{{{e}^{2x}}-1}{2x}-5\lim\limits_{x\to 0}\,\dfrac{{{e}^{5x}}-1}{5x} \\ =2-5=-3 \)