Given that the tangent line to the graph of $\displaystyle f(x)$ at (0,0) has equation $\displaystyle 2y=x$ and that $\displaystyle f(x)$ has a horizontal asymptote at $\displaystyle \infty$ with equation $\displaystyle y=2$, find the value of

$\displaystyle \displaystyle\lim_{x\to{0^{+}}} {\frac{sin(2x)}{f(x)}} - x^2f(\frac{1}{x})$