Infinity - infinity to the power of infinity

$\begin{aligned} L & = \lim_{x \to \infty} \left(\sqrt{x^2+2x+3} - \sqrt{x^2+3} \right)^x \\ & = \lim_{x \to \infty} \exp \left(\ln\left(\sqrt{x^2+2x+3} - \sqrt{x^2+3} \right)^x\right) \\ & = \exp \left(\lim_{x \to \infty} x\ln\left(\sqrt{x^2+2x+3} - \sqrt{x^2+3} \right)\right) \\ & = \exp \left(\lim_{x \to \infty} \frac {\ln\left(\frac {2x}{\sqrt{x^2+2x+3} + \sqrt{x^2+3}}\right)}{\frac 1x} \right) \quad \quad \small \blue{\text{A 0/0 case, L'Hôpital's rule applies.}} \\ & = \small \exp \left(\lim_{x \to \infty} \frac {\sqrt{x^2+2x+3} + \sqrt{x^2+3}}{2x} \left(\frac {2\sqrt{x^2+2x+3} + 2\sqrt{x^2+3}-2x\left(\frac {x+1}{\sqrt{x^2+2x+3}} + \frac x{\sqrt{x^2+3}}\right)}{(\sqrt{x^2+2x+3} + \sqrt{x^2+3})^2} \right) (-x^2)\right) \\ & = \small \exp \left(\lim_{x \to \infty} - \frac 12 \left(\sqrt{x^2+2x+3} + \sqrt{x^2+3} - \frac {x^2+x}{\sqrt{x^2+2x+3}} - \frac {x^2}{\sqrt{x^2 +3}}\right) \right) \\ & = \small \exp \left(\lim_{x \to \infty} -\frac 12 \left(\frac {(x+3)\sqrt{x^2+3}+3\sqrt{x^2+2x+3}}{\sqrt{(x^2+2x+3)(x^2+3)}}\right) \right) \\ & = \small \exp \left(\lim_{x \to \infty} -\frac 12 \left(\frac {\left(1+\frac 3x\right)\sqrt{1+\frac 3{x^2}}+\frac 3{x^2} \sqrt{x^2+2x+3}}{\sqrt{\left(1+\frac 2x+\frac 3{x^2}\right) \left(1+\frac 3{x^2} \right)}}\right) \right) \\ & = e^{-\frac 12} = \boxed{\frac 1{\sqrt e}} \end{aligned}$

Reference: L'Hôpital's rule

Substituting $x$ by $\dfrac{1}{h}$ we reduce the given expression as

$\left (\dfrac{\sqrt {1+2h+3h^2}-\sqrt {1+3h^2}}{h}\right) ^{\frac{1}{h}}\approx (1-\frac{h}{2})^{\frac{1}{h}}$

As $x$ approaches $\infty,h$ approaches $0$ . Hence the required limit is $e^{\displaystyle \lim_{h \to 0} \dfrac{\ln (1-\frac{h}{2})}{h}}=e^{-\frac{1}{2}}=\dfrac{1}{\sqrt e}$ .