A calculus problem by Jose Sacramento

$\begin{aligned} L & = \lim_{x \to 0} \frac {\ln((2e)^{x+1}-e)-1}{e^x-1} & \small \color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \frac {\color{#3D99F6}\frac d{dx} (\ln((2e)^{x+1}-e)-1)}{\frac d{dx} (e^x-1)} & \small \color{#3D99F6} \text{See note.} \\ & = \lim_{x \to 0} \frac {\color{#3D99F6}\frac {\ln(2e)(2e)^{x+1}} {(2e)^{x+1}-e}}{e^x} \\ & = 2 \ln (2e) \\ & = \ln 4 + 2 \end{aligned}$

$\implies a+b = 4+2 = \boxed{6}$

---

Note:

$\small \begin{aligned} f'(x) & = \frac d{dx} (\ln {\color{#3D99F6}((2e)^{x+1}-e)}-1) & \color{#3D99F6} \text{Let }u = (2e)^{x+1}-e \\ & = \frac d{dx} (\ln {\color{#3D99F6}u}-1) \\ & = \frac 1u \cdot \frac {du}{dx} \\ & = \frac 1u \cdot \frac d{dx} \left((2e)^{x+1}-e\right) \\ & = \frac {\ln(2e)(2e)^{x+1}} {(2e)^{x+1}-e} \end{aligned}$