Directional derivative: what is the relation between definition by limit and definition as dot product? I'm trying to get intuition about why gradient is pointing to the direction of the steepest ascent. I got confused because I found that directional derivative is explained with help of gradient and gradient is explained with help of directional derivative.
Please explain what are the exact steps that lead from 
directional derivative defined by the limit $\nabla_{v} f(x_0) = \lim_{h\to 0} \frac{f(x_0+hv)-f(x_0)}h$ to directional derivative defined as dot product of gradient and vector $\nabla_{v} f(x_0) = \nabla f(x_0)\cdot{v}$ ?
In other words how to prove the following? $$\lim_{h\to 0} \frac{f(x_0+hv)-f(x_0)}h = \nabla f(x_0)\cdot{v}$$

Let $s$ be a distance $u=\left<u_1,u_2,\ldots,u_n \right>$  a unit vector, $P(a_1,a_2,\ldots,a_n)$ a point and and $f(x_1,x_2,\ldots, x_n)$ be a real valued function differentiable in an open neighborhood $D$ of $P$  
Then parameterize a line distance $s$ away from $P$ in the direction of $u$
$$
\begin{split}
x_1=&a_1+su_1\\
x_2=&a_2+su_2\\
& \vdots \\
x_n=&a_n+su_n\\
\end{split}
$$
Then apply the chain rule to find $\frac{df}{ds}$ along this line at least in the open neighborhood $D$ the following applies
$$
\begin{split}
\frac{df}{ds}&=\frac{\partial f}{\partial x_1}\frac{d x_1}{d s}
+\frac{\partial f}{\partial x_2}\frac{d x_2}{d s}+\cdots+
\frac{\partial f}{\partial x_n}\frac{d x_n}{d s}\\
&=\frac{\partial f}{\partial x_1}u_1
+\frac{\partial f}{\partial x_2}u_2+\cdots+
\frac{\partial f}{\partial x_n}u_n\\
&=\left<\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\ldots,\frac{\partial f}{\partial x_n}
\right>\cdot 
\left<u_1,u_2,\ldots,u_n
\right>\\
&= \nabla f \cdot u
\end{split}
$$
You can evaluate all the functions involved at $P$ to get 
$$
\frac{df}{ds} |_P= \nabla f(P) \cdot u = \left|\nabla f(P)\right| \cos \theta
$$
where $\theta$ is the angle between $\nabla f$ and $u$. The biggest possible value for the directional derivative happens when  $\cos \theta=1$ (this is the biggest possible value for cosine) which is when $\theta=0$. That is, $\nabla f$ and $u$ have to be in the same direction. Therefore, $\nabla f$ is in the direction of greatest increase of $f$. 
