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

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up vote 1 down vote accepted

Taking ln of both sides, you obtain, for $x\to+\infty$: \begin{equation} 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. \end{equation} Thus your limit is $\sqrt e$.

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