# Calculate the limit of logarithm functions

How can I calculate limit expressed by: $$\lim_{x\to \infty} \left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln(x)}$$

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Taking ln of both sides, you obtain, for $x\to+\infty$: $$x\ln(x)\ln\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)\sim x\ln(x)\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}-1\right)=\frac{x\ln(x)}{\ln(x^2+2x+3)}\ln\left(\frac{x^2+3x+4}{x^2+2x+3}\right)\sim \frac{x\ln(x)}{\ln(x^2+2x+3)}\left(\frac{x^2+3x+4}{x^2+2x+3}-1\right)= \frac{x\ln(x)}{\ln(x^2+2x+3)}\left(\frac{x+1}{x^2+2x+3}\right)\sim \frac{\ln(x)}{\ln(x^2+2x+3)}\sim \frac{\ln(x)}{\ln(x^2)}\to\frac 12.$$ Thus your limit is $\sqrt e$.