题目内容
10.用洛必达法则求下列极限:(1)$\underset{lim}{x→0}$$\frac{1-cosx}{{x}^{2}}$
(2)$\underset{lim}{x→0}$$\frac{{e}^{x}-{e}^{-x}-2x}{x-sinx}$
(3)$\underset{lim}{x→{0}^{+}}\frac{lnsin3x}{lnsinx}$
(4)$\underset{lim}{x→0}(\frac{1}{x}-\frac{1}{{e}^{x}-1})$.
分析 (1)由洛必达法则化简$\underset{lim}{x→0}$$\frac{1-cosx}{{x}^{2}}$=$\underset{lim}{x→0}$$\frac{sinx}{2x}$=$\underset{lim}{x→0}$$\frac{cosx}{2}$=$\frac{1}{2}$;
(2)由洛必达法则化简$\underset{lim}{x→0}$$\frac{{e}^{x}-{e}^{-x}-2x}{x-sinx}$=$\underset{lim}{x→0}$$\frac{{e}^{x}+{e}^{-x}-2}{1-cosx}$=$\underset{lim}{x→0}$$\frac{{e}^{x}-{e}^{-x}}{sinx}$=$\underset{lim}{x→0}$$\frac{{e}^{x}+{e}^{-x}}{cosx}$=2;
(3)由洛必达法则化简$\underset{lim}{x→{0}^{+}}\frac{lnsin3x}{lnsinx}$=$\underset{lim}{x→{0}^{+}}$$\frac{3\frac{1}{sin3x}cos3x}{\frac{1}{sinx}cosx}$=$\underset{lim}{x→{0}^{+}}$$\frac{3sinxcos3x}{sin3xcosx}$=$\underset{lim}{x→{0}^{+}}$$\frac{3sinx}{sin3x}$=$\underset{lim}{x→{0}^{+}}$$\frac{3cosx}{3cos3x}$=1;
(4)由洛必达法则化简$\underset{lim}{x→0}(\frac{1}{x}-\frac{1}{{e}^{x}-1})$=$\underset{lim}{x→0}$$\frac{{e}^{x}-x-1}{x({e}^{x}-1)}$=$\underset{lim}{x→0}$$\frac{{e}^{x}-1}{{e}^{x}-1+x{e}^{x}}$=$\underset{lim}{x→0}$$\frac{{e}^{x}}{{e}^{x}+{e}^{x}+{xe}^{x}}$=$\frac{1}{2}$.
解答 解:(1)$\underset{lim}{x→0}$$\frac{1-cosx}{{x}^{2}}$=$\underset{lim}{x→0}$$\frac{sinx}{2x}$=$\underset{lim}{x→0}$$\frac{cosx}{2}$=$\frac{1}{2}$;
(2)$\underset{lim}{x→0}$$\frac{{e}^{x}-{e}^{-x}-2x}{x-sinx}$=$\underset{lim}{x→0}$$\frac{{e}^{x}+{e}^{-x}-2}{1-cosx}$
=$\underset{lim}{x→0}$$\frac{{e}^{x}-{e}^{-x}}{sinx}$=$\underset{lim}{x→0}$$\frac{{e}^{x}+{e}^{-x}}{cosx}$=2;
(3)$\underset{lim}{x→{0}^{+}}\frac{lnsin3x}{lnsinx}$=$\underset{lim}{x→{0}^{+}}$$\frac{3\frac{1}{sin3x}cos3x}{\frac{1}{sinx}cosx}$=$\underset{lim}{x→{0}^{+}}$$\frac{3sinxcos3x}{sin3xcosx}$
=$\underset{lim}{x→{0}^{+}}$$\frac{3sinx}{sin3x}$=$\underset{lim}{x→{0}^{+}}$$\frac{3cosx}{3cos3x}$=1;
(4)$\underset{lim}{x→0}(\frac{1}{x}-\frac{1}{{e}^{x}-1})$=$\underset{lim}{x→0}$$\frac{{e}^{x}-x-1}{x({e}^{x}-1)}$
=$\underset{lim}{x→0}$$\frac{{e}^{x}-1}{{e}^{x}-1+x{e}^{x}}$=$\underset{lim}{x→0}$$\frac{{e}^{x}}{{e}^{x}+{e}^{x}+{xe}^{x}}$=$\frac{1}{2}$.
点评 本题考查了洛必达法则的应用.
A. | R | B. | Φ | C. | {0} | D. | {x|x≠0} |
A. | f(x)在区间(0,+∞)上是减函数 | B. | f(x)在区间(0,+∞)上是增函数 | ||
C. | f(x)在区间(0,+∞)上先增后减 | D. | f(x)在区间(0,+∞)上是先减后增 |
A. | 无穷间断点 | B. | 可去间断点 | C. | 连续点 | D. | 震荡间断点 |