Define $f_{\lambda,\sigma}(x):[0,1]\to\mathbb{R}$ as $$f_{\lambda,\sigma}(x)= \Theta(x-\lambda)|x-\lambda|^{\sigma},$$ where $\Theta$ is the Heaviside Function and $\lambda,\sigma\in (0,1)$. Then here, on page 8, it is claimed that
$$\lim_{t\to 0}\frac{|f_{\lambda,\sigma}(\lambda+t)-f_{\lambda,\sigma}(\lambda-t)|}{|2t|^{\sigma}}=\frac{1}{2^{\sigma}}.$$
How did they evualuate this limit? Substituting $t=0$, we get $\frac{0}{0}$. So I've tried noticing that $$f_{\lambda,\sigma}'(\lambda-t) = \lim_{t\to 0}\frac{f_{\lambda,\sigma}(\lambda+t)-f_{\lambda,\sigma}(\lambda-t)}{2t}.$$
Working out the derivative of $f_{\lambda,\sigma}$, I get $$f_{\lambda,\sigma}'(x)=\delta(x-\lambda)|x-\lambda|^{\sigma}+\Theta(x-\lambda)\sigma|x-\lambda|^{\sigma-1}.$$
Substituting above, $$\lim_{t\to 0}\frac{|f_{\lambda,\sigma}(\lambda+t)-f_{\lambda,\sigma}(\lambda-t)|}{|2t|^{\sigma}}=\lim_{t\to 0}\frac{|f_{\lambda,\sigma}(\lambda+t)-f_{\lambda,\sigma}(\lambda-t)|}{|2t|}\frac{1}{|2t|^{\sigma}}=\frac{\sigma |t|^{\sigma-1}}{|2t|^{\sigma-1}}=\frac{\sigma}{2^{\sigma-1}}.$$
Where have I gone wrong?
Assume $t\to 0^+$. Since $f_{\lambda,\sigma}(x) = 0$ if $x< \lambda$, the second term $f_{\lambda,\sigma}(\lambda - t) = 0$. So $$ \lim_{t\to 0^+} \frac{|f_{\lambda,\sigma}(\lambda + t) - f_{\lambda,\sigma}(\lambda - t)|}{|2t|^\sigma} = \lim_{t\to 0^+}\frac{f_{\lambda,\sigma}(\lambda + t)}{(2 t)^\sigma}= \lim_{t\to 0^+}\frac{t^\sigma}{2^\sigma t^\sigma} = \frac{1}{2^\sigma} $$ If $t \to 0^-$, the first term is zero but the absolute value of the difference is the same.
Actually, if you just check the cases, I think you'll notice that $$ |f_{\lambda,\sigma}(\lambda + t) - f_{\lambda,\sigma}(\lambda - t)| = |t|^\sigma $$ for all $t$. So the limit is straightforward.
I am not sure how your derivative argument is supposed to work (are you aiming to apply L'Hôpital's rule?) but I see you wrote $|2t|^\sigma = |2t| \cdot |2t|^\sigma$, and you should probably have a $\sigma - 1$ power there.