题目内容
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}}$.
分析 (1)$\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.
点评 本题考查了洛必达法则的应用.
A. | [-4,4] | B. | (-4,4) | C. | [-4,0)∪(0,4] | D. | (-∞,4)∪(4,+∞) |
A. | [-4,2] | B. | [-4,-1)∪(-1,2] | C. | (-4,2) | D. | (-4,-1)∪(-1,2) |