Everything But the Kitchen Sink

First look at the limit. Letting the integral be $F(x)$ and $f(x) = x$ , we have both of these functions going to $0$ as $x \rightarrow 0$ , and thus we have a limit of the indeterminate form $\frac{0}{0}.$ As both functions satisfy the conditions required to apply L'Hopital's rule, we apply this rule (and use the Fundamental Theorem of Calculus) to find that our limit is equal to

$S = \lim_{x \rightarrow 0} (e^{2x}\cos(3x) + e^{3x}\sinh(2x))^{\frac{1}{x}},$

which is now of the indeterminate form $1^{\infty}.$ So now we need to look at what happens to $\ln(S)$ in the limit as $x \rightarrow 0.$ We then end up with

$\lim_{x \rightarrow 0} \ln(S) = \lim_{x \rightarrow 0} \dfrac{\ln(e^{2x}\cos(3x) + e^{3x}\sinh(2x))}{x}$ ,

which is of the indeterminate form $\frac{0}{0}$ , so apply L'Hopital's to get

$\lim_{x \rightarrow 0} \dfrac{2e^{2x}\cos(3x) - 3e^{2x}\sin(3x) + 3e^{3x}\sinh(2x) + 2e^{3x}\cosh(2x)}{e^{2x}\cos(3x) + e^{3x}\sinh(2x)} = 4.$

But since we must take the natural log anyway we just have $\boxed{4}$ as our final answer.