Help me prove the result of this limit: $\lim_{x\to \infty} {\log_{x^3+1} (x^2+1) \over \log_{2e^x+1} (x^2+5x)}$ The limit is 
$$\lim_{x\to \infty} {\log_{x^3+1} (x^2+1) \over \log_{2e^x+1} (x^2+5x)}$$
I know that it should be equal to  $\infty$ but i have yet to prove it. Please help me do so.

Hint 1: $\displaystyle \log_a b = \frac{\ln b}{\ln a}$. So $$\frac{\log_{x^3+1} (x^2+1)}{\log_{2e^x+1} (x^2+5x)} = \frac{\frac{\ln (x^2+1)}{\ln (x^3+1)}}{\frac{\ln (x^2+5x)}{\ln (2e^x+1)}} = \frac{\ln (x^2+1) \ln (2e^x+1)}{\ln (x^3+1) \ln (x^2+5x)}$$
Hint 2: For very large $x$, $x^2+1 \approx x^2$ and $x^3 + 1 \approx x^3$. So our equation becomes $$\frac{\ln (x^2+1) \ln (2e^x+1)}{\ln (x^3+1) \ln (x^2+5x)} \approx \frac{\ln (x^2) \ln (2e^x+1)}{\ln (x^3) \ln (x^2+5x)} \approx \frac{2\ln (x) \ln (2e^x+1)}{3\ln (x) \ln (x^2+5x)} \approx \frac{2}{3} \frac{\ln (2e^x+1)}{\ln (x^2+5x)}$$
Can you take it from here now?
