The tangent line is the best "linear" approximation to the graph of a differentiable function I wanted to understand what it means that the tangent line is the best linear approximation to the graph of a differentiable function at the point of tangency.
I've looked in several books and I don't understand anything yet.
P.S. Where I wrote "linear approximation" read "affine approximation"... I saw that they don’t use this much, but I thought about like the parabola and the tangent at the point (0,0) which is the x axis... Like: How do I prove it is the best approximation? It's intuitively obvious, but I want a proof.
Thanks in advance to whoever help me. ^^
 A: Let $f$ be a function defined on an open subset $U$ of $\mathbb{R}$, differentiable at $a\in U$. Then for $x\in U$
$$f(x)=f(a)+f'(a)(x-a)+o_a(x-a) \tag{1}$$
What you want to show is that the function $x\mapsto f'(a)(x-a)+f(a)$ is the only affine function verifying this kind of property. To do so, suppose that there is another affine function $g(x)=\alpha (x-a)+\beta$ defined on $\mathbb{R}$ such that
$$\forall x\in U, f(x)=g(x)+o_a(x-a) \tag{2}$$
Evaluating $(2)$ in $a$ yields
$$\beta=g(a)=f(a)$$
Moreover, on the one hand the definition of $g$ gives
$$\forall x\in\mathbb{R}, x\neq a, \frac{g(x)-g(a)}{x-a}=\frac{\alpha(x-a)}{x-a}=\alpha \tag{3}$$
And on the other hand, equation $(2)$ implies that
$$\forall x\in U, x\neq a, \frac{g(x)-g(a)}{x-a}=\frac{f(x)-f(a)}{x-a}+o_a(1)$$
Letting $x\rightarrow a$ in this expression and using $(3)$ gives
$$\alpha = f'(a)$$
Therefore, $g(x)=f'(a)(x-a)+f(a)$. We have shown that $x\mapsto f'(a)(x-a)+f(a)$ is the affine function verifying $(1)$.
A: First, you should think of "linear" as "first order". Then consider your function $f$, and let's approximate it near some point $x_0$ in the domain with a linear function $L(x) = a(x-x_0)+b$. You want:

*

*zeroth order approximation: $\lim\limits_{x\to x_0} (f(x) - (a(x-x_0)+b))= 0$.

*first order approximation: $\lim\limits_{x \to x_0} \dfrac{f(x)-(a(x-x_0)+b)}{x-x_0} = 0$.

The zeroth order requirement gives that $b = f(x_0)$. Plug that into the first order requirement to get that $a = f'(x_0)$. So the linear approximation you want is $$L(x) = f(x_0) + f'(x_0)(x-x_0).$$Play around with second order approximations and so on to deduce higher order Taylor polynomials.
