a) \mathop {\lim }\limits_{x \to 0}\frac{tanx-sinx}{x^3}=\mathop {\lim }\limits_{x \to 0}\frac{sinx-sinx.cosx}{x^3}=\mathop {\lim }\limits_{x \to 0}\frac{sinx(1-cosx)}{x^3}=$\mathop {\lim }\limits_{x \to 0}\frac{sinx.sin^2(\frac{x}{2})}{x^3}=\mathop {\lim }\limits_{x \to 0}\frac{x.\frac{sinx}{x}.(\frac{x}{2})^2.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}=\mathop {\lim }\limits_{x \to 0}\frac{x^3\frac{sinx}{x}.\frac{1}{4}.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}=\frac{1}{4}$
a)
\mathop {\lim }\limits_{x \to 0}\frac{tanx-sinx}{x^3}=$\mathop {\lim }\limits_{x \to 0}\frac{
\frac{sinx-sinx.cosx
}{cosx}}{x^3}
=\mathop {\lim }\limits_{x \to 0}\frac{sinx(
\frac{1-cosx
}{cosx})}{x^3}
=\mathop {\lim }\limits_{x \to 0}\frac{sinx.
(\frac{sin^2\frac{x}{2}
}{cosx})}{x^3}
=\mathop {\lim }\limits_{x \to 0}\frac{x.\frac{sinx}{x}.
\frac{1}{cosx}(\frac{x}{2})^2.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}
=\mathop {\lim }\limits_{x \to 0}\frac{x^3\frac{sinx}{x}.\frac{1}{4
cosx}.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}
=\frac{1}{4cosx}=\frac{1}{4}$