\Rightarrow \displaystyle \lim_{x \to +\infty} {\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln{x}}}= e^{\diplaystyle \lim_{x \to +\infty}{\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}-1\right)x\ln{x}}}=