Let $f \colon \mathbb R^d \to \mathbb R \cup \{ \infty \}$ be proper and lower semicontinuous. Let $x, u \in (\mathbb R^d, | \cdot |_2)$ with $0 < | u |_2 \le 1$. Choose sequences $(t_n)_{n \in \mathbb N} \subset \mathbb{R}_{> 0}$ and $(v_n)_{n \in \mathbb N} \subset \mathbb R^d$ such that $$ \lim_{n \to \infty} \frac{f(x) - f(x + t_n v_n)}{t_n} = \liminf_{\substack{t \searrow 0 \\ v \to u}} \frac{f(x) - f(x + t v)}{t}. $$ The limit on the left side exists.
My question Does the limit \begin{equation} \tag{$\star$} \label{eq:1} \lim_{n \to \infty} \frac{f(x) - f(x + t_n v_n)}{t_n | v_n |_2} \end{equation} exist, too?
Ideas. The following limit exists: \begin{equation} \tag{$\star \star$} \label{eq:2} \lim_{n \to \infty} \frac{f(x) - f(x + t_n v_n)}{t_n | u |_2} = \lim_{n \to \infty} \frac{f(x) - f(x + t_n v_n)}{t_n | v_n |_2} \frac{| v_n |_2}{| u |_2}. \end{equation} We know that $\frac{| v_n |_2}{| u |_2} \to 1$ and that the product of the limits is the limit of the product if both limits exists. Since $| v_n | \to | u |$, for every $\varepsilon > 0$, there exists a $N \in \mathbb N$ such that $| | v_n |_2 - | u |_2 | < \varepsilon$ for all $n > N$. Hence for every $\varepsilon > 0$ we have $$ \lim_{n \to \infty} \frac{f(x) - f(x + t_n v_n)}{t_n (| u |_2 + \varepsilon)} \le \lim_{n \to \infty} \frac{f(x) - f(x + t_n v_n)}{t_n | v_n |_2} \le \lim_{n \to \infty} \frac{f(x) - f(x + t_n v_n)}{t_n (| u |_2 - \varepsilon)}. $$ Both the outer left and the outer right limit exist and for $\varepsilon \searrow 0$ converge to the middle one. Is this enough to conclude that \eqref{eq:1} exists and is equal to \eqref{eq:2}?
Your last inequalities must be reversed if the numerator is $<0.$
Anyway, if $\lim_{n\to\infty}\frac{f(x)-f(x+t_nv_n)}{t_n}=L,$ simply write $$\lim_{n\to\infty}\frac{f(x)-f(x+t_nv_n)}{t_n|v_n|_2}=\frac L{\lim_{n\to \infty}|v_n |_2}=\frac L{|u|_2}.$$