L'Hospital's Rule

Let $\mathbb{T}$ be a time scale. Let $\overline{\mathbb{T}} = \mathbb{T} \cup \left\{\sup \mathbb{T} \right\} \cup \left\{ \inf \mathbb{T} \right\}$. If $\infty \in \overline{\mathbb{T}},$ then we call $\infty$ left-dense and if $-\infty \in \overline{\mathbb{T}}$ we call $-\infty$ right-dense. Define for left-dense $t_0 \in \mathbb{T}$ and $\epsilon > 0$, the set $$L_{\epsilon}(t_0) := \left\{ t \in \mathbb{T} \colon 0 < t_0-t < \epsilon \right\}.$$ is non-empty by definition of left-dense. If $\infty \in \overline{\mathbb{T}},$ then $$L_{\epsilon}(\infty) := \left\{t \in \mathbb{T} \colon t > \dfrac{1}{\epsilon} \right\}$$ is clearly nonempty.

Let $h \colon \mathbb{T} \rightarrow \mathbb{R}$. Define the notation for left-dense $t_0 \in \overline{\mathbb{T}}$, $$\liminf_{t \rightarrow t_0^-} h(t) := \displaystyle\lim_{\epsilon \rightarrow 0^+} \inf_{t \in L_{\epsilon}(t_0)} h(t).$$ Define a similar $R_{\epsilon}$ function for right-dense points.

Theorem (L'Hospital's Rule 1): Assume $f,g$ $\Delta$-differentiable on $\mathbb{T}$ and for some left-dense $t_0 \in \overline{\mathbb{T}}$, $$\displaystyle\lim_{t \rightarrow t_0^-} f(t) = \lim_{t \rightarrow t_0^-} g(t)=0.$$ Further suppose there is an $\epsilon > 0$ such that for all $t \in L_{\epsilon}(t_0),$ $$g(t)>0;g^{\Delta}(t)<0.$$ Then $$\displaystyle\liminf_{t \rightarrow t_0^-} \dfrac{f^{\Delta}(t)}{g^{\Delta}(t)} \leq \liminf_{t \rightarrow t_0^-} \dfrac{f(t)}{g(t)} \leq \limsup_{t \rightarrow t_0^-} \dfrac{f^{\Delta}(t)}{g^{\Delta}(t)}.$$

Proof: █

Theorem (L'Hospital's Rule 2): Assume $f,g$ are $\Delta$-differentiable functions on $\mathbb{T}$ and for some left-dense $t_0 \in \overline{\mathbb{T}}$, $$\displaystyle\lim_{t \rightarrow t_0^-} g(t) = \infty.$$ Also suppose there exists an $\epsilon > 0$ such that for all $t \in L_{\epsilon}(t_0)$, $$g(t)>0;g^{\Delta}(t)>0.$$ Then $$\displaystyle\lim_{t \rightarrow t_0^-} \dfrac{f^{\Delta}(t)}{g^{\Delta}(t)} = r \in \overline{\mathbb{R}}$$ implies $$\displaystyle\lim_{t \rightarrow t_0-} \dfrac{f(t)}{g(t)}=r.$$

Proof: █