Can You Solve This Question Without Using L'Hopital's Rule?

$\displaystyle = \lim_{x \to 0} \frac{g\left(g\left(x\right)\right)-g\left(e\right)}{x}$

$\displaystyle = \lim_{x \to 0} \frac{g\left(g\left(x\right)\right)-g\left(g\left(0\right)\right)}{x-0}$

$\displaystyle = \lim_{x \to 0} \frac{g\left(g\left(x\right)\right)-g\left(g\left(0\right)\right)}{g\left(x\right)-g\left(0\right)}\cdot\frac{g\left(x\right)-g\left(0\right)}{x-0}$

When $x \to 0$ , $g(x) \to g(0) = e$

$\displaystyle = \lim_{g(x) \to e} \left(\frac{g\left(g\left(x\right)\right)-g\left(e\right)}{g\left(x\right)-e}\right)\cdot \lim_{x \to 0} \left(\frac{g\left(x\right)-g\left(0\right)}{x-0}\right)$

$\displaystyle = g'\left(e\right)\cdot g'\left(0\right)$

$\displaystyle = \boxed{4e^{2e+2}}$

By the definition of the derivative, $\begin{aligned} \lim_{x\to0} \frac{g(g(x)) - g(e)}{x} &= \lim_{x\to0}\frac{(g\circ g)(x) - (g\circ g)(0)}{x-0} \\ &= (g\circ g)'(0) \\ &= g'(g(0))\cdot g'(0) \\ &= (2e^{2e+1})\cdot (2e) \\ &= \boxed{4e^{2e+2}} \end{aligned}$