A question of an example on Spivak Calclulus On Manifolds I understand almost all the steps except one in the following example about differentiation on Spivak's Calculus on Manifolds:

We shall later discover a simple way of finding $Df(a)$. For the moment let us consider the function $f:\mathbf{R}^2\to\mathbf{R}$ defined by $f(x,y)=\sin x$. Then $Df(a,b)=\lambda$ satisfies $\lambda(x,y)=(\cos a)\cdot x$. To prove this, note that
$$
\lim_{(h,k)\to 0}\frac{|f(a+h,b+k)-f(a,b)-\lambda(h,k)|}{|(h,k)|}=\lim_{(h,k)\to 0}\frac{|\sin(a+h)-\sin a-(\cos a)\cdot h|}{|(h,k)|}.
$$
Since $\sin '(a)=\cos a$, we have
$$
\lim_{h\to 0}\frac{|\sin(a+h)-\sin a-(\cos a)\cdot h|}{|h|}=0.
$$
Since $|(h,k)|\geqslant |h|$, it is also true that $$\lim_{h\to 0}\frac{|\sin(a+h)-\sin a-(\cos a)\cdot h|}{|(h,k)|}=0.$$

Indeed, my doubt is that it is shown that the first limit is less than or equal than $0$. But Spivak's just say that it's equal to zero. Why is that? Probably is a simple thing that I'm missing, but I don't see it.
 A: So, I'm going to try and give as complete of an answer as I can here. I interpret your question as asking why having $$\lim_{(h,k) \to 0} \frac{|\sin(a+h) - \sin(a) - (\cos(a)) \cdot h|}{|(h,k)|} \leq 0$$ is enough to deduce the limit is $0$ exactly. I'll try to prove the following result:

Proposition: Let $A$ be an open subset of $\mathbb{R}^n$ and suppose we have $f: A \to \mathbb{R}$ be a non-negative function (so $f(x) \geq 0$ for all $x \in A$ ). Let $x_0$ be a limit point of $A$. If $\lim_{x \to x_0} f(x) \leq 0$, then the limit is $0$ exactly.

Proof: Fix $\epsilon > 0$, and let $L = \lim_{x \to x_0} f(x)$. By the definition of the limit, there exists a $\delta > 0$ such that if $0 < ||x-x_0||< \delta$, then $|f(x) - L| < \epsilon$. We know this will imply $f(x) < L + \epsilon \leq \epsilon$ because $L \leq 0$ by hypothesis. Furthermore, since $f$ is a non-negative function, $-\epsilon < f(x)$. Putting  everything together, we have the compound inequality $-\epsilon < f(x) < \epsilon$ for $0 < ||x-x_0|| < \delta$. But this just says that $|f(x) - 0| < \epsilon$, so we've shown $0$ is a limit. Since limits of real-valued functions are unique, when they exist, we automatically must have $L = 0$.
Going back to the first limit I presented in this answer, define $f: \mathbb{R^2} \setminus \{(0,0)\} \to \mathbb{R}$ given by $$f(x,y) = \frac{|\sin(a+x) - \sin(a) - (\cos(a)) \cdot y|}{|(x,y)|}$$
The function $f$ outlined above satisfies all of the criteria for the proposition I just proved. That is, $f$ is a non-negative function (because of the norms), we can let $A = \mathbb{R}^2 \setminus \{(0,0)\}$ (this is an open subset of
$\mathbb{R}^2$), and $x_0 = (0,0)$ is a limit point of $A$.
