If P denotes the number of points of intersection of ${\mathrm{y}}^2=7\mathrm{x} \mathrm{and} {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}+2=0$, then value of $\underset{\mathrm{x}\to \mathrm{p}}{\lim }\frac{\mathrm{x}\sin \left(\mathrm{sinx}\right)-{\sin }^2\mathrm{x}}{{\mathrm{x}}^6}, \mathrm{is}$

$$\begin{gathered} \mathrm{L}=\underset{\mathrm{x}\to 0}{lim} \frac{{\sin }^2\mathrm{x}}{{\mathrm{x}}^2}\left[\frac{\mathrm{xsin}(\sin \mathrm{x})}{{\sin }^2\mathrm{x}}-1\right]\frac{1}{{\mathrm{x}}^4} \\ \begin{array}{l}=\underset{\mathrm{x}\to 0}{lim} \frac{1}{{\mathrm{x}}^4}\left[\frac{\mathrm{x}}{\sin \mathrm{x}}\left\{1-\frac{{\sin }^2\mathrm{x}}{6}+\frac{{\sin }^4\mathrm{x}}{120}-1\right\}\right]\\ =\underset{\mathrm{x}\to 0}{lim} \frac{1}{{\mathrm{x}}^4}\frac{\left(1-\frac{{\sin }^2\mathrm{x}}{6}+\frac{{\sin }^4\mathrm{x}}{120}\dots ..\left(1-\frac{{\mathrm{x}}^2}{6}+\frac{{\mathrm{x}}^4}{120}\dots .\right)\right)}{1-\frac{{\mathrm{x}}^2}{6}+\dots .}\\ =\underset{\mathrm{x}\to 0}{lim} \frac{1}{{\mathrm{x}}^4}\left[-\frac{{\mathrm{x}}^2}{6}{\left(1-\frac{{\mathrm{x}}^2}{6}+\dots \right)}^2+\frac{{\mathrm{x}}^4}{120}+\frac{{\mathrm{x}}^6}{6}-\frac{{\mathrm{x}}^4}{120}+..\right]\\ \left.=\underset{\mathrm{x}\to 0}{lim} \frac{1}{{\mathrm{x}}^4}\left[\frac{{\mathrm{x}}^4}{18}+\dots .\right]=\frac{1}{18}\right]\end{array} \end{gathered}$$