\displaystyle \lim_{x \to +\infty} {\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}}=\lim_{x \to +\infty} {\frac{\ln{x^2}+\ln(1+\frac{3x+4}{x^2})}{\ln{x^2}+\ln(1+\frac{2x+3}{x^2})}}=\lim_{x \to +\infty} {\frac{1+\frac{\ln(1+\frac{3x+4}{x^2})}{\ln{x^2}}}{1+\frac{\ln(1+\frac{2x+3}{x^2})}{\ln{x^2}}}}=1