On negation of lipschiz continuity Let $f: [a,b] \to R$ be continuous function which is not Lipschitz continuous.
Can we say there exist $x \in [a,b] $ and  strictly monotone sequences, $\{x_n\}_{n=1}^{\infty} \subseteq [a,b] $ and $\lambda_{n} \in \mathbb{R^+} $ such that $x_n \to x$ and $\lambda_{n}  \to + \infty $,  
$$|f(x_n) - f(x)| > \lambda_{n}  |x_n - x|  $$ 
for all $n \in \mathbb{N}$? 
P.S: Cleary we don't need to care about monotonicity of those two sequences! 

Consider the function
$$
f(x) = 
\begin{cases}
x \sin (1/x), &\text{if}\ x\neq 0,\\
0, & \text{if}\ x = 0.
\end{cases}
$$
This function is continuous in $\mathbb{R}$, it is not Lipschitz continuous, and it has continuous derivative in $\mathbb{R}\setminus\{0\}$, hence around each point $x\neq 0$ it is locally Lipschitz continuous.
Hence the only candidate for the point $x$ in your claim is $x=0$.
On the other hand
$$
|f(y) - f(0)| \leq |y| \qquad \forall y.
$$
