Sequential definition of differentiation 
Let $f$ be a function continuous and derivable at $x $. Let ${u_n}$ and $v_n$ be two sequence such that   ${u_n}\leq x \leq {v_n}$. Show that $ |u_n - v_n |$ tends to zero implies  $$\frac{f(u_n)-f(v_n)}{u_n-v_n}=f'(x)$$

What I have tried is $ |u_n - v_n |$ tends to zero implies $ {v_n} ,  {u_n} $ converges  to $x$ since $f$ being continuous at $x$. We can use sequential criterion. From that how can we conclude that $$\frac{f(u_n)-f(v_n)}{u_n-v_n}=f'(x)$$
 A: Assuming $\forall  n, u_n<x<v_n$, write $$\displaystyle \frac{f(u_n)-f(v_n)}{u_n-v_n}= \frac{f(v_n)-f(x)}{v_n-x}\frac{v_n-x}{v_n-u_n} + \frac{f(x)-f(u_n)}{x-u_n}\frac{x-u_n}{v_n-u_n}$$
Note that $\frac{v_n-x}{v_n-u_n}=O(1)$ and $\frac{x-u_n}{v_n-u_n}=O(1)$, hence $$\displaystyle \frac{f(u_n)-f(v_n)}{u_n-v_n}=(f'(x)+o(1))\frac{v_n-x}{v_n-u_n} + (f'(x)+o(1))\frac{x-u_n}{v_n-u_n}=f'(x)+o(1)O(1)+o(1)O(1)$$
Hence $$\displaystyle \frac{f(u_n)-f(v_n)}{u_n-v_n}=f'(x) + o(1)$$
A: I don't know what Sequential Definition of Derivative you are using. But to be clear let me state the version of the definition that I will be assuming.

Sequential Definition of Derivative. A function $f:D(\subseteq\mathbb{R})\to\mathbb{R}$ is said to be differentiable at a point $x\in D$ if for all sequence $(x_n)_{n\in\mathbb{N}}$ converging to $x$ with $x_n\ne x$ for all $x$ the limit, $$\displaystyle\lim_{n\to\infty}
\dfrac{f(x_n)-f(x)}{x_n-x}$$exists.

Observe that assuming this definition we can immediately say that $u_n<x<v_n$. Now, \begin{align}\left\lvert\dfrac{f(u_n)-f(v_n)}{u_n-v_n}-f'(x)\right\rvert&=\left\lvert\dfrac{f(u_n)-f(v_n)-(u_n-v_n)f'(x)}{u_n-v_n}\right\rvert\\&=\left\lvert\dfrac{(f(u_n)-f(x))-(f(v_n)-f(x))-((u_n-x)-(v_n-x))f'(x)}{u_n-v_n}\right\rvert\\&=\left\lvert\dfrac{(f(u_n)-f(x)-(u_n-x)f'(x))+(f(x)-f(v_n)-(x-v_n)f'(x))}{u_n-v_n}\right\rvert\\&=\left\lvert\left(\dfrac{f(u_n)-f(x)-(u_n-x)f'(x)}{u_n-v_n}\right)+\left(\dfrac{f(x)-f(v_n)-(x-v_n)f'(x)}{u_n-v_n}\right)\right\rvert\\&\le\left\lvert\dfrac{f(u_n)-f(x)-(u_n-x)f'(x)}{u_n-v_n}\right\rvert +\left\lvert\dfrac{f(x)-f(v_n)-(x-v_n)f'(x)}{u_n-v_n}\right\rvert\\&=\left\lvert\dfrac{f(u_n)-f(x)-(u_n-x)f'(x)}{u_n-x}\right\rvert\color{red}{\left\lvert\dfrac{u_n-x}{u_n-v_n}\right\rvert }+\left\lvert\dfrac{f(x)-f(v_n)-(x-v_n)f'(x)}{x-v_n}\right\rvert \color{red}{\left\lvert\dfrac{x-v_n}{u_n-v_n}\right\rvert}\\&\le\left\lvert\dfrac{f(u_n)-f(x)-(u_n-x)f'(x)}{u_n-x}\right\rvert+\left\lvert\dfrac{f(x)-f(v_n)-(x-v_n)f'(x)}{x-v_n}\right\rvert\\&=\left\lvert\dfrac{f(u_n)-f(x)}{u_n-x}-f'(x)\right\rvert+\left\lvert\dfrac{f(x)-f(v_n)}{x-v_n}-f'(x)\right\rvert\end{align} 
Can you justify how we got from the "red colored" step to the next step? Can you take it from here?
A: The methods offered so far handle the case (i) $u_n<x<v_n$ with $v_n-u_n\to 0$ only, whereas a complete solution would handle the weaker case (ii) that $u_n\leq x \leq v_n$ with $0<v_n-u_n$ and  $v_n-u_n\to 0$.  Since  there is a more trivial way to prove this anyway I will add it.
The fact that $f'(x)$ exists, by definition, means that for any $\epsilon>0$ there is a $\delta>0$ so that whenever $x-\delta<u\leq x \leq v< x+\delta$ one has
$$-\epsilon(x-u) \leq f(x)-f(u) -f'(x)(x-u) \leq \epsilon(x-u)$$
and
$$-\epsilon(v-x) \leq f(v)-f(x) -f'(x)(v-x) \leq \epsilon(v-x).$$
Now merely add:
$$-\epsilon(v-u) \leq f(v)-f(u) -f'(x)(v-u) \leq \epsilon(v-u).$$
To complete, using assumption (ii), choose $N$ so that $v_n-u_n<\delta $ if $n\geq N$ and obtain
$$\left| \frac{f(v_n)-f(u_n)}{v_n-u_n} -f'(x)\right| \leq \epsilon $$
for all $n\geq N$.

This also demonstrates (the nonsequential version) that the ordinary derivative $f'(x)$ can be equivalently defined as the limit as $(u,v)\to (x,x)$ with  the restrictions that $u\leq x \leq v$ and $u\not=v$ of the limit $$ \lim \frac{f(v)-f(u)}{v-u}$$
sometimes called the straddled derivative.  If you neglect to insist that $u$ and $v$ straddle $x$ then this would be a much stronger statement than the existence of the ordinary derivative.  See "strong derivative" or "unstraddled derivative" or "Peano strict derivative."
