Answer Check: Show $L_1 = L_2$ Guys I need an answer check:
Let $L_1$ and $L_2$ be two linear mappings from $R \rightarrow R^n$ safisfiying:
$\lim_{h\to0} \frac{f(a+h)-f(a)-L(h)}{h} = 0$
prove that $L_1 = L_2$.
Ok this is what I got:
$\lim_{h\to0} \frac{f(a+h)-f(a)-L_1(h)}{h} = 0$
$\lim_{h\to0} \frac{f(a+h)-f(a)-L_2(h)}{h} = 0$
so
$\lim_{h\to0} \frac{f(a+h)-f(a)-L_2(h)}{h} = \lim_{h\to0} \frac{f(a+h)-f(a)-L_1(h)}{h}$
thus
$\frac{f(a+h)-f(a)-L_1(h)}{h} = \frac{f(a+h)-f(a)-L_2(h)}{h}$
thus
$L_1(h) = L_2(h)$
thus 
$L_1 = L_2$
 A: We have
$$\lim_{h\to0} \frac{f(a+h)-f(a)-L_2(h)}{h} = \lim_{h\to0} \frac{f(a+h)-f(a)-L_1(h)}{h}$$
so
$$\lim_{h\to0} \frac{L_1(h)-L_2(h)}{h} = 0,$$
and by linearity of $L_1$ and $L_2$ we find $L_1(1)=L_2(1)$.
Now again by linearity we have:
$$\forall x\in\mathbb{R},\quad L_1(x)=xL_1(1)=xL_2(1)=L_2(x),$$
hence $L_1=L_2.$
A: Terribly wrong! We cannot conclude anything like $a_n=b_n$ from $\lim(a_n)=\lim(b_n)$, it's nonsense!
Instead, use that a linear mapping $\Bbb R\to\Bbb R^n$ must look like $h\mapsto h\cdot v$ for some fixed $v\in\Bbb R^n$ (and varying $h\in\Bbb R$). So, we get that $L_i(h)=h\cdot v_i$ for some vectors $v_1$ and $v_2$. Then, try to prove that the hypothesis is quivalent to
$$\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=v_i\,.$$
A: I'll do a more general case where $f,L_j:\mathbb{R}^m\longrightarrow\mathbb{R}^n$. This is even true from a normed vector space to another normed vector space. Now the assumptions are:
$$
\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)-L_j(h)}{\|h\|}=0.
$$
Note that taking the norm corresponds to $|h|$ in your case, adn it follows from your hypothsesis. Subtracting these assumptions for $L_1$ and $L_2$, this amounts to showing that if $L=L_1-L_2$ is linear
$$
\lim_{h\rightarrow0}\frac{L(h)}{\|h\|}=0\qquad\Rightarrow\qquad L=0.
$$
Fix $h_0\neq 0$. Then for $t>0$ tendind to $0$, we have $h=th_0\longrightarrow 0$  so
$$
0=\lim_{t\rightarrow 0}\frac{L(th_0)}{\|th_0\|}=\lim_{t\rightarrow 0}\frac{tL(h_0)}{t\|h_0\|}=\frac{L(h_0)}{\|h_0\|}.
$$
Hence $L(h_0)=0$. And since $h_0\neq 0$ was arbitrary, it follows that $L=0$. The fact that $L(0)=0$ is automatic by linearity.
Note: what this says inparticular is that if $f$ is differentiable at $a$, then the derivative is unique.
