若条件符合,洛必达法则可连续复用,直至求出极限为止
first exercise
\[\lim_{x \to 0} \frac{x-sinx}{x^{3}}=? \]\[\\ \\ \]\[\lim_{x \to 0} \frac{x-sinx}{x^{3}}=\lim_{x \to 0} \frac{(x)'-(sinx)'}{(x^{3})'}=\lim_{x \to 0} \frac{1-cosx}{3x^{2}} \]\[\\ \\ \]\[\lim_{x \to 0} \frac{1-cosx}{3x^{2}}=\lim_{x \to 0} \frac{(1)'-(cosx)'}{(3x^{2})'}=\lim_{x \to 0} \frac{0-(-sinx)}{6\cdot (x^{2})'} \]\[\\ \\ \]\[=\lim_{x \to 0} \frac{sinx}{6x} =\frac{1}{6} \cdot \lim_{x \to 0}\frac{sinx}{x} =\frac{1}{6} \]\[\\ \\ \]\[\therefore \lim_{x \to 0} \frac{x-sinx}{x^{3}}=\frac{1}{6} \]
标签：洛必达,3x,frac,训练,sinx,lim,法则,cosx From： https://www.cnblogs.com/Preparing/p/16757383.html