Notations on differential

Question

Is all this the same?

$$ \frac{di(0^{+})}{dt} = \left( \frac{di(t)}{dt} \right)_{t=0^{+}} = \lim_{t\rightarrow 0^{+}} \frac{di(t)}{dt} = \lim_{t\rightarrow 0 \over t > 0} \frac{di(t)}{dt} $$

EDIT: Im sorry, I just realize that I wrote it wrong. It should be:

$$ \frac{di(0^{+})}{dt} = \left( \frac{di(t)}{dt} \right)_{\scriptsize t=0^{+}} = \left( \frac{di(t)}{dt} \right)_{\overset{t=0}{\scriptsize t>0}} = \lim_{\scriptsize t\rightarrow 0^{+}} \frac{\Delta i(t)}{\Delta t} = \lim_{\overset{t\rightarrow 0}{\scriptsize t > 0}} \frac{\Delta i(t)}{\Delta t} = \lim_{\scriptsize t\rightarrow 0^{+}} \frac{i(t + \Delta t)}{\Delta t} = \lim_{\overset{t\rightarrow 0}{\scriptsize t > 0}} \frac{i(t + \Delta t)}{\Delta t} $$

Answer 1

You need to add some conditions to make this correct. Denote $f'(a^+)=\lim_{x\to a^+}f'(x) $ and $f'_+(a)=\lim_{x\to a^+}\frac{f(x)-f(a)}{x-a}$. We have the following property:

If $f$ is differentiable in $(a,b)$ and $f'(a^+)$ exists, then $f'_{+}(a)$ exists and $f'_{+}(a)=f'(a^+)$.

We can prove it by mean value theorem. For $\Delta x>0$, we have $$ f'_{+}(a)=\lim_{\Delta x\to 0^+}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\to 0^+}\frac{f(a+\Delta x)-f(a)}{\Delta x}=\lim_{\Delta x\to 0^+}\frac{f'(a+\theta \Delta x)\Delta x}{\Delta x}=f'(a^+). $$