$\begin{align}
& \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x-\sin x\cos x}{{{\tan }^{3}}ax}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x(1-\cos x)}{{{\tan }^{3}}ax}=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin x}{\tan \,ax}\times \frac{2{{\sin }^{2}}\frac{x}{2}}{{{\tan }^{2}}ax} \right) \\
& =\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{\tan ax}\times \underset{x\to 0}{\mathop{\lim }}\,2{{\left( \frac{\sin \frac{x}{2}}{\tan ax} \right)}^{2}}=\frac{1}{a}\times 2\times {{\left( \frac{\frac{1}{2}}{a} \right)}^{2}}=\frac{1}{a}\times 2\times \frac{1}{4{{a}^{2}}}=\frac{1}{2{{a}^{3}}} \\
& \Rightarrow \frac{1}{2{{a}^{3}}}=4\Rightarrow {{a}^{3}}=\frac{1}{8}\Rightarrow a=\frac{1}{2} \\
\end{align}$
عباس ابراهیمی علویجه 
