A question from Spivak text

Question

I am having trouble understanding the following argument from Spivak's Calculus on Manifolds:

The above argument is clear.

But I cannot figure out how for any fixed $k,$ $$\lim_{h\to0}\dfrac{|\sin(a+h)-\sin a-(\cos a).h|}{|(h,k)|}=0$$ implies that $$\lim_{(h,k)\to0}\dfrac{|\sin(a+h)-\sin a-(\cos a).h|}{|(h,k)|}=0.$$ Please help!

Added: Please see if it is correct: for $\epsilon>0~\exists~\delta>0$ such that $\frac{|\sin(a+h)−\sin(a)−\cos(a)h|}{|(h,k)|}<\epsilon$ for $|h|<\delta$ ($k$ is fixed). Then $|(h,k)|<\delta\implies|h|<\delta\implies\frac{|\sin(a+h)−\sin(a)−\cos(a)h|}{|(h,k)|}<\epsilon.$

Answer 1

We have $|(h,k)|\ge |h|$, then $$ 0\le\left|\frac{f(h)}{(h,k)}\right|=\left|\frac{f(h)}{h}\frac{h}{(h,k)}\right|\le \left|\frac{f(h)}{h}\right|. $$ Now $(h,k)\to 0$ $\Rightarrow$ $h\to 0$. What can you say about the limit?

Answer 2

What is in the numerator is always of the order of $h^2$, : $$\sin(a+h)-\sin(a)-\cos(a)h=\mathcal{O}(h^2)$$ while $|(h,k)|$ is at least $|(h,k)|=\mathcal{O}(h)$ if $k\sim h$.

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