I'm not any sort of an expert at limits, but this looks

so I'm going to take a shot at it.

First since $\displaystyle \lim_{x \to +\infty}f(x) = 2$ we know that $\displaystyle \lim_{x \to 0^+}f \left ( \frac{1}{x} \right ) = 2$

Thus

$\displaystyle \lim_{x \to 0+}x^2 f \left ( \frac{1}{x} \right ) = \lim_{x \to 0^+} 2x^2 = 0$

Next we have $\displaystyle \lim_{x \to 0^+} \frac{sin(2x)}{f(x)}$. We know that $\displaystyle \lim_{x \to 0}f(x) = 0$ since the tangent line to y = f(x) at x = 0 is 2y = x. Thus the limit is of the form $\displaystyle \frac{0}{0}$ and we may apply L'Hopital's rule:

$\displaystyle \lim_{x \to 0^+} \frac{sin(2x)}{f(x)} = \lim_{x \to 0^+} \frac{2 cos(2x)}{\frac{1}{2}} = \lim_{x \to 0^+}4cos(2x) = 4$

(Because $\displaystyle f(x) \approx \frac{1}{2}x$ near x = 0.)

Thus

$\displaystyle \lim_{x \to 0^+} { \frac{sin(2x)}{f(x)}} - x^2f(\frac{1}{x}) = 4 - 0 = 4$

-Dan