I'm self studying mathematics for interest and I've come across a proof, which on the face of it should be straightforward but I have a slight issue. The statement is as follows:
If a vector function $\mathbf{f}$ is differentiable at $\mathbf{a}$, then $\mathbf{f}$ is continuous at $\mathbf{a}$.
Definition Suppose that $\mathbf{a}\in \mathbb{R}^n$ and $V$ is an open set containing $\mathbf{a}$ and $\mathbf{f}:V\rightarrow \mathbb{R}^m$. The function $\mathbf{f}$ is said to be differentiable at $\mathbf{a}$ if and only if there is a $\mathbf{T} \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$ such that the function \begin{equation} \varepsilon(\mathbf{h}) := \mathbf{f}(\mathbf{a}+\mathbf{h})-\mathbf{f}(\mathbf{h})-\mathbf{T}(\mathbf{h}) \end{equation} (defined for $\|\mathbf{h}\|$ sufficiently small) satisfies \begin{equation} \frac{\varepsilon(\mathbf{h})}{\|h\|}\rightarrow \mathbf{0}\;\;\text{as}\;\;\mathbf{h}\rightarrow \mathbf{0} \end{equation}
The proof is fairly straightforward. It's the beginning of the proof I'm slightly unsure of.
Proof Suppose that $\mathbf{f}$ is differentiable at $\mathbf{a}$. By the definition above there exists a $\mathbf{T} \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$ and a $\delta > 0$ such that \begin{equation} \|\mathbf{f}(\mathbf{a}+\mathbf{h})-\mathbf{f}(\mathbf{h})-\mathbf{T}(\mathbf{h})\| \leq \|\mathbf{h}\|\;\;\;\;\;(*) \end{equation} for all $\|\mathbf{h}\| < \delta$. Then, using the triangle inequality and by the definition of the operator norm we have \begin{equation} \|\mathbf{f}(\mathbf{a}+\mathbf{h})-\mathbf{f}(\mathbf{h})\| \leq \|\mathbf{T}\| \|\mathbf{h}\|+\|\mathbf{h}\| \end{equation} Taking the limit as $\mathbf{h} \rightarrow \mathbf{0}$, using the fact that $\|\mathbf{T}\|$ is a finite real number we conclude that $\mathbf{f}(\mathbf{a}+\mathbf{h})\rightarrow \mathbf{f}(\mathbf{a})$.
I don't understand where the weak inequality comes from in $(*)$. To my mind, as $\frac{\varepsilon(\mathbf{h})}{\|h\|}\rightarrow \mathbf{0}$ we have that for all $\epsilon >0$ there exists a $\delta > 0$ such that $\|\mathbf{h}\|< \delta$ implies. \begin{equation} \left\|\frac{\mathbf{f}(\mathbf{a}+\mathbf{h})-\mathbf{f}(\mathbf{h})-\mathbf{T}(\mathbf{h})-\mathbf{0}}{\|\mathbf{h}\|} \right\|<\epsilon \end{equation} Taking $\epsilon =1$ and rearranging I get \begin{equation} \|\mathbf{f}(\mathbf{a}+\mathbf{h})-\mathbf{f}(\mathbf{h})-\mathbf{T}(\mathbf{h})\| < \|\mathbf{h}\| \end{equation} then the rest of the proof is the same. Where does the weak inequality come from? Am I missing something silly? Obviously I'm aware that $< \implies \leq$ but I don't see the necessity of the weak inequality for the proof. By taking the limit of the strict inequality I get a weak inequality anyway. I'm assuming I'm misunderstanding something. Any help would be greatly appreciated.