Find the matrix representative of a linear transformation $T∈L(\mathbb R,\mathbb R^2).$ Let $f(x) =(x^2, \sin x)$. Find the matrix representative of a linear transformation $T∈L(\mathbb R,\mathbb R^2)$ which satisfies $\lim_{h→0}\frac{||f(x+h)−f(x)−T(h)||}{h}= 0.$
The norm part of this problem is throwing me off. I believe I have the right answer for the problem, its just the matrix of the derivatives of $f(x).$ I just don't know how to explain why.
 A: Start from scratch and calculate:
$f(x+h)-f(x)=((x+h)^2-x^2, \sin (x+h)-\sin x)\Rightarrow $
$f(x+h)-f(x)-T(x)h=((x+h)^2-x^2-T_1(x)h, \sin (x+h)-\sin x-T_2(x)h)\Rightarrow $
$\frac{1}{h}\left(f(x+h)-f(x)-T(x)h\right)=\left(\frac{(x+h)^2-x^2-T_1(x)}{h}, \frac{\sin (x+h)-\sin x-T_2(x)}{h}\right).$
Now, if we choose $T_1(x)=2x$ and $T_2(x)=\cos x$, the limit in your question is zero, as required. And of course, $T_1(x)$ and $T_2(x)$ are the derivatives of $f_1(x)=x^2$ and $f_2(x)=\sin x$.
The matrix representation of $T(x)$ is then $\begin{pmatrix}
2x\\ 
\cos x
\end{pmatrix}$ and it operates on the "vector" $t\in \mathbb R$ by $\begin{pmatrix}
2x\\ 
\cos x
\end{pmatrix}t= \begin{pmatrix}
2xt\\ 
(\cos x)t
\end{pmatrix}.$
Since the domain vector space is just $\mathbb R,$ we can take $t=1$ to recover the definition of the derivative usually seen in elementary calculus courses.
A: Since $T\in L(\mathbb R,\mathbb R^2)$, it is of the form
$$
T_x(y)=(a_xy,c_xy).
$$
for certain $a_x,c_x\in\mathbb R$.
So, for some fixed $x$, we want
$$
\lim_{h\to0}\frac{\Big\|\big((x+h)^2-x^2-a_xh,\sin(x+h)-\sin x-c_xh\big)\Big\|}{h}=0.
$$
The quotient expands to (using the MVT)
$$
\frac{\sqrt{\big((2r_h-a_x)h+h^2\big)^2+\big(h\,(-c_x+\cos s_h)\big)^2}}{h},
$$
where $r_h,s_h$ are between $x$ and $x+h$. Considering the case $h>0$ (the case $h<0$ is similar) the square quotient (which has to go to zero) can be written as
$$
\Big((2r_h-a_x)+h\Big)^2+\Big(-c_x+\cos s_h\Big)^2.
$$
When $h\to0$ it becomes
$$
(2x-a_x)^2+(-c_x+\cos x)^2. 
$$
So, to be $0$, we need $a_x=2x$, $c_x=\cos x$. Then
$$
T_xh=(2xh,(\cos x) h). 
$$
