Motivation for the differentiability of a function $f: D \subseteq \Bbb R^m \rightarrow \mathbb R$ Suppose a function $f: D \subseteq \Bbb R^m \rightarrow \mathbb R$ is defined on an open subset $D \subseteq \Bbb R^m$. Then, we define that : $f$ is differentiable at $p$ if there exists a linear function $L: \Bbb R^m \rightarrow \Bbb R$ and a function $\eta: D_p \subseteq \Bbb R^m \rightarrow \Bbb R$ such that :
$$f(p+h)-f(p)=L(h)+ ||h|| \eta (h)$$
where $h \in D_p, ~~\lim_{||h|| \rightarrow 0 }\eta(h)=0.$
$D_p$ is defined as $=\{h \in \Bbb R^m:p+h \in D\}$

I wanted to understand the motivation behind this definition. Here's what I could think:
$\lim_{h \rightarrow 0} \dfrac{f(p+h)-f(p)}{||h||}$ represents the derivative at $p$
$\dfrac{f(p+h)-f(p)}{||h||}$ can be approximated as $f'(p)+\eta(h)$ where $\eta(h)$ represents an error function.
Thus, $f(p+h)-f(p)= ||h||f'(p) + ||h|| \eta(h)$.
Now, our definition calls $||h|| f'(p)$ as a linear function $L(h)$.
If we can prove that $||h|| f'(p)$ is linear, then this argument is a sufficient motivation for this definition of differentiability of function $f$.
But: for $\lambda \in \Bbb R: ||h_1+\lambda h_2|| f'(p) \ne ||h_1||f'(p) + |\lambda|~||h_2||f'(p)$ and hence not linear !!
If it would have been, there would have been a perfect motivation for the definition. Now, in  such a case, why do we call $L(h)$  a linear function in the original definition?
Could someone tell me how did this definition come about. What is the motivation behind this definition? Thanks!
You said:


$\lim_{h \rightarrow 0} \dfrac{f(p+h)-f(p)}{||h||}$ represents the derivative at $p$


No, it doesn't. Next, you write:


$\dfrac{f(p+h)-f(p)}{||h||}$ can be approximated as $f'(p)+\eta(h)$ where $\eta(h)$ represents an error function.


Again, this is false.

The definition of differentiability is that there exist a linear function $L:\Bbb{R}^m\to \Bbb{R}$ such that
\begin{align}
f(p+h) - f(p) &amp;= L(h) + \lVert h\rVert\eta(h) \tag{$\ddot{\smile}$}
\end{align}
where $\lim_{h\to 0} \eta(h) = 0$. Typically, the notation used is that $L := Df_p$. Also, if we define $\Delta f_p(h) := f(p+h) - f(p)$, then the above equation becomes very memorable:
\begin{align}
\Delta f_p(h) &amp;= Df_p(h) + \lVert h\rVert \eta(h).
\end{align}
This is exactly the formal way of saying that differentiable functions are locally approximately linear, because it says the actual change in the function (at the point $p$ by an amount $h$) $\Delta f_p(h)$ is equal to a linear part $Df_p(h)$ plus an error term $\lVert h\rVert\eta(h)$, and this error term is &quot;small&quot; in the sense $\eta(h)\to 0$ as $h\to 0$.
So, to address (1) above, it is $L= Df_p$ which is the derivative at $p$ (by definition).
For (2), we have
\begin{align}
\dfrac{f(p+h) - f(p)}{\lVert h \rVert} &amp;= \dfrac{Df_p(h)}{\lVert h\rVert} + \eta(h) \\
&amp;= Df_p\left(\dfrac{h}{\lVert h \rVert}\right) + \eta(h)
\end{align}
so, you can interpret this however you want. But the point remains: $L(\cdot) = Df_p(\cdot)$ is by definition a linear transformation which approximates changes in $f$ (i.e which is approximately equal to $\Delta f_p(\cdot)$).

In the comments you ask:

But, if we were to go by the basic definition of a derivative: it should be $f′(p)\cdot \lVert h\rVert$.

What do you mean by basic definition? The definition you wrote in the question is the definition of derivative in multivariable calculus. Do you mean the definition of single-variable calculus? If that's what you meant then the thing is you need to learn to re-interpret the definition in single variable calculus. We are often taught in the case of $f:\Bbb{R}\to \Bbb{R}$ to think of $f'(p)$ geometrically as &quot;the instantaneous slope at $p$&quot;, because we define (if the limit exists)
\begin{align}
f'(p):= \lim_{h\to 0}\dfrac{f(p+h) - f(p)}{h}
\end{align}
so of course, geometrically this forces us to think in terms of slopes.
What I'm now suggesting to you is to think in terms of &quot;local linear approximations (this allows for a much easier transition to multivariable calculus) and to rewrite this definition as
\begin{align}
\lim_{h\to 0}\dfrac{f(p+h) - f(p) - f'(p)\cdot h}{h} &amp;= 0 \tag{$*$}
\end{align}
In this case, the mapping $\Bbb{R}\to \Bbb{R}$, $h\mapsto f'(p)\cdot h$ is a linear transformation which approximates the actual change $\Delta f_p(h):= f(p+h) - f(p)$.
Note that in this case, we are able to divide by $h$ because it is a real number and not a vector. But notice that $(*)$ is entirely equivalent to $(**)$:
\begin{align}
\lim_{h\to 0} \dfrac{|f(p+h) - f(p) - f'(p)\cdot h|}{|h|} &amp;= 0 \tag{$**$}
\end{align}
and in this form, the relation with definition of differentiability in higher dimensions is much more clear, because $(\ddot{\smile})$ is entirely equivalent to the following statement (with appropriate domains and target spaces):

There exists a linear transformation $L$ such that
\begin{align}
\lim_{h\to 0}\dfrac{\lVert f(p+h) - f(p) - L(h)\rVert}{\lVert h\rVert} &amp;= 0.
\end{align}
The motivation is this:
You want to know how $f(x$) changes with few movements of $x$ arround some open set $V$ with $x$ inside it. The task can be reached with a estimation of $f(x) \equiv T(x)$ where $T(x)$ is a linear function.
If this is true, then you can say: &quot;oh well, I wanted to know what happened with $x$ moving inside the open set $V$, I can do this by $x+tu$, where $t$ is a tiny number and $v$ a small direction vector, so If I examine $$T(x+tu)=T(x)+tT(u).$$ My goal is to find the best $T(x)$ for $f(x)$, so I look for $T(x)$ from last equation:
$$T(x) = T(x+tu) - tT(u)$$
If $t$ is small as possible, then I'm taking the best linear function approximation of $f$ trough $x$ point.
You can identify some elements with respect the derivative definiton, such as the factor $t$ which is equivalent to your $||h||$, both must to be small enought for the good aproximation.