Question

16．求极限$\underset{lim}{x→0}$$\frac{{∫}_{0}^{x}sintdt}{xtanx}$．

Answer 1

分析 求定积分${∫}_{0}^{x}$sintdt=1-cosx，从而利用洛比达法则求极限$\underset{lim}{x→0}$$\frac{1-cosx}{xtanx}$即可．

解答 解：${∫}_{0}^{x}$sintdt=-cost$|\left.\begin{array}{l}{x}\\{0}\end{array}\right.$=1-cosx，
故$\underset{lim}{x→0}$$\frac{{∫}_{0}^{x}sintdt}{xtanx}$
=$\underset{lim}{x→0}$$\frac{1-cosx}{xtanx}$
=$\underset{lim}{x→0}$$\frac{cosx（1-cosx）}{xsinx}$
=$\underset{lim}{x→0}$$\frac{1-cosx}{xsinx}$
=$\underset{lim}{x→0}$$\frac{sinx}{sinx+xcosx}$
=$\underset{lim}{x→0}$$\frac{cosx}{cosx+cosx-xsinx}$=$\frac{1}{2}$．

点评 本题考查了定积分的求法及洛比达法则的应用．