Proof that the derivative is unique?

Given a subset $\Omega$ of $\mathbb{R}^n$ and a function $\sigma: \Omega \to \mathbb{R}^n$, we define its derivative at $x$ to be a linear operator $\sigma'$ such that $$ \lim_{y \to 0} \frac{||\,\sigma(x + y) - \sigma(y) - \sigma'(y)\,||}{||\,y\,||} = 0 $$

If this operator exists (i.e. if $\sigma$ is differentiable at $x$) then how do we prove that $\sigma'$ is unique?

(Sorry about the unconventional operator notation, it's just that I don't like how $\sigma'y $ would have looked).

Suppose there are linear operators $\sigma_{1}'$ and $\sigma_{2}'$ such that $$ \lim_{\|y\|\rightarrow 0}\frac{\|\sigma(x+y)-\sigma(x)-\sigma_{j}'(x)\|}{\|y\|} =0, \;\;\; j =1,2. $$ Because $\sigma_{2}'$ and $\sigma_{1}'$ are linear, $$ (\sigma_{2}'-\sigma_{1}')(y) = \{\sigma(x+y)-\sigma(x)-\sigma_{1}'(y)\}-\{\sigma(x+y)-\sigma(x)-\sigma_{2}'(y)\} $$ Applying the triangle inequality gives $$ \lim_{\|y\|\rightarrow 0}\frac{\|(\sigma_{1}'-\sigma_{2}')(y)\|}{\|y\|}=0 $$ In particular, for any non-zero vector $y$, one has $\alpha y \rightarrow 0$ as the scalar $\alpha$ tends to 0. Therefore, by linearity of $\sigma_{1}'-\sigma_{2}'$, $$ 0= \lim_{\alpha\rightarrow 0}\frac{\|(\sigma_{1}'-\sigma_{2}')(\alpha y)\|}{\|\alpha y\|} = \lim_{\alpha\rightarrow 0}\frac{\|(\sigma_{1}'-\sigma_{2}')(y)\|}{\|y\|} = \frac{\|(\sigma_{1}'-\sigma_{2}')(y)\|}{\|y\|}. $$ So $\|(\sigma_{2}'-\sigma_{1}')(y)\|=0$ for all non-zero vectors $y$, which implies that $\sigma_{2}'=\sigma_{1}'$.