Showing $\limsup_{h \to {0}}\frac{O(h^2)}{h^2}<\infty$

Question

Let

$$y(h)=1-2\sin^{2}(2\pi h) , f(y)=\frac{2}{1+\sqrt(1-y^2)} $$

Justify the statement

$$f(y(h))=2-4\sqrt{2}\pi+O(h^2)$$

where

$$\limsup_{h \to {0}}\frac{O(h^2)}{h^2}<\infty$$

Answer 1

By definition

$$g(x)=\mathcal{O}(f(x))\iff\exists M>0,\; |g(x)|\le M|f(x)|.$$

Hence

$$\limsup {\mathcal{O}(f(x))\over f(x)}\le \limsup \left|{\mathcal{O}(f(x))\over f(x)}\right|\le M$$

taking $f(h)=h^2$ gives the result.