Limited

We have that

$\displaystyle\lim_{x \to 0} \sin(10x)=0 \land \lim_{x \to 0} \sin(7x)=0$

and

$\displaystyle\lim_{x \to 0} \frac{\sin(10x)'}{ \sin(7x)'}=\displaystyle\lim_{x \to 0} \frac{10\cos(10x)}{ 7\cos(7x)}=\displaystyle\frac{10}{7}$

So, by L'Hôpital's rule, we have that

$\displaystyle\lim_{x \to 0} \frac{\sin(10x)}{ \sin(7x)}=\displaystyle\lim_{x \to 0} \frac{\sin(10x)'}{ \sin(7x)'}=\displaystyle\frac{10}{7}$ .

$\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \frac{{\sin (10x)}}{{\sin (7x)}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin (10x)}}{x}}}{{\frac{{\sin (7x)}}{x}}} = \mathop {\lim }\limits_{x \to 0} \frac{{10 \times \frac{{\sin (10x)}}{{10x}}}}{{7 \times \frac{{\sin (7x)}}{{7x}}}}\mathop = \limits_{(1)} \frac{{10}}{7}\\ \\ (1)\quad \mathop {\lim }\limits_{x \to 0} \frac{{\sin (ax)}}{{ax}} = 1 \end{array}$