Question

17．（1）$\underset{lim}{x→0}$$\frac{\sqrt{1+xsinx}-1}{{e}^{3x}-1}$
（2）$\underset{lim}{x→0}$$\frac{tanx-sinx}{x（arcsinx）^{2}}$．

Answer 1

分析 （1）$\underset{lim}{x→0}$$\frac{\sqrt{1+xsinx}-1}{{e}^{3x}-1}$=$\underset{lim}{x→0}$$\frac{\frac{sinx+xcosx}{2\sqrt{1+xsinx}}}{3{e}^{3x}}$=0，
（2）$\underset{lim}{x→0}$$\frac{tanx-sinx}{x（arcsinx）^{2}}$=$\underset{lim}{x→0}$$\frac{1-co{s}^{3}x}{（arcsinx）^{2}+2xarcsinx}$=$\underset{lim}{x→0}$$\frac{3sinx}{4arcsinx+2x}$=2．

解答 解：（1）$\underset{lim}{x→0}$$\frac{\sqrt{1+xsinx}-1}{{e}^{3x}-1}$
=$\underset{lim}{x→0}$$\frac{\frac{sinx+xcosx}{2\sqrt{1+xsinx}}}{3{e}^{3x}}$=0，
（2）$\underset{lim}{x→0}$$\frac{tanx-sinx}{x（arcsinx）^{2}}$
=$\underset{lim}{x→0}$$\frac{\frac{1}{co{s}^{2}x}-cosx}{（arcsinx）^{2}+2xarcsinx•\frac{1}{\sqrt{1-{x}^{2}}}}$
=$\underset{lim}{x→0}$$\frac{1-co{s}^{3}x}{（arcsinx）^{2}+2xarcsinx}$
=$\underset{lim}{x→0}$$\frac{3co{s}^{2}xsinx}{2arcsinx\frac{1}{\sqrt{1-{x}^{2}}}+2arcsinx+2x\frac{1}{\sqrt{1-{x}^{2}}}}$
=$\underset{lim}{x→0}$$\frac{3sinx}{4arcsinx+2x}$
=$\underset{lim}{x→0}$$\frac{3cosx}{\frac{4}{\sqrt{1-{x}^{2}}}+2}$=$\frac{3}{6}$=2．

點(diǎn)評(píng) 本題考查了洛必達(dá)法則的應(yīng)用．