Uniqueness of a Tangent Linear Map Let $E$ and $F$ be Banach spaces and let $X \subset E$ be open. Call maps $\phi,\psi:X\rightarrow F\;$  tangent
at the point $p \in X$ if
$$
\lim_{x \to p} = \frac{|\phi(x)-\psi(x)|_F}{|x-p|_E}=0
$$
For a given map $f:X\rightarrow F$, I am trying to show that there is at most one bounded linear map $L:E\rightarrow F$ that is tangent to
the map $g_L(x) = f(p) + L(x - p)$. Demonstrating this claim will show that the derivative is unique. 
So, suppose there exists two such maps $L_1$ and $L_2$. Applying the triangle inequality
and properties of the operator norm of bounded linear functions,
$$
0  = \lim_{x \to p}\frac{1}{|x - p|}\left( |f(x) - f(p) - L_1(x - p)| - |f(x) - f(p) - L_2(x - p)|   \right)
$$
$$
\leq \lim_{x \to p}\frac{1}{|x - p|}\left( |f(x) - f(p)| + |L_1(x-p)| - |f(x) - f(p)| - |L_2(x-p)|  \right)
$$
$$
 = \lim_{x \to p}\frac{1}{|x - p|} \left(|L_1(x-p)| - |L_2(x-p)|\right)
$$
$$
 \leq \lim_{x \to p}\frac{1}{|x - p|} \left(|L_1|\cdot |x-p| - |L_2|\cdot |x-p|\right)
$$
$$
= |L_1| - |L_2|
$$
So,
$$
0 \leq |L_1| - |L_2| \implies |L_1| \geq |L_2|.
$$
By a symmetrical argument $|L_2| \geq |L_1|$ and therefore, $|L_1| = |L_2|$.
So, my question is, am I on the right track? I believe my proof correctly shows that
$|L_1| = |L_2|$, however, this does not imply that $L_1 = L_2$. Can this approach be salvaged? 
 A: The first inequality is not justified, because $-|f(x)-f(p)-L_2(x-p)|\geq -|f(x)-f(p)|-|L_2(x-p)|$ rather than the other way around.  Also, there is no justification for the limits in the second and third lines to exist.  (In particular, note that $\lim\limits_{x\to p} \frac{|L_1(x-p)|}{|x-p|}$ usually does not exist for a linear map $L_1$.)  And if the limit in the third line does exist, the subsequent inequality isn't justified, because $-|L_2(x-p)|\geq -|L_2||x-p|$ rather than the other way around.
Rather, the following approach could be used. For each $x\neq p$, the triangle inequality implies that 
$$\begin{align*}
\frac{|(L_1-L_2)(x-p)|}{|x-p|}&=\frac{|L_1(x-p)-L_2(x-p)|}{|x-p|}\\
&\leq \frac{|f(x)-f(p)-L_1(x-p)|}{|x-p|}+\frac{|f(x)-f(p)-L_2(x-p)|}{|x-p|}.
\end{align*}$$  Since the hypothesis is that each of the fractions on the right-hand side goes to $0$ as $x\to p$, it follows that for all $\varepsilon>0$, there exists $\delta>0$ such that $|(L_1-L_2)(x-p)|\leq \varepsilon |x-p|$ whenever $0<|x-p|<\delta$.  Therefore $|L_1-L_2|=\sup\limits_{0<|y|<\delta}\frac{|(L_1-L_2)(y)|}{|y|}\leq \varepsilon$.  Since $\varepsilon>0$ was arbitrary, this shows that $L_1=L_2$.
A: In fact, we can copy the proof of the uniqueness of the differential of a function. Fix $h\neq 0$. Then we should have 
$$\lim_{t\to 0^+}\frac{|f(p)+L(th+p)-L_j(th+p)|}{t|h|}=0\quad \forall j\in\{1,2\},$$
so $\lim_{t\to 0^+}\frac{L_1(th+p)-L_2(th+p)}{t|h|}=0$ and so 
$$\lim_{t\to 0^+}\frac{(L_1-L_2)h+(L_1-L_2)\left(\frac pt\right)}{|h|}=0,$$
hence denoting $L'=L_1-L_2$: $\lim_{t\to 0^+}\frac 1t|L'(p)|=|L'(h)|$ for all $h$, hence $L'(p)=0$ and $L'(h)=0$. Since $h$ was arbitrary, $L_1=L_2$. 
