Solve $\lim_{x \rightarrow 0} \frac{\ln (x+1)}{\ln (2x+1)}$ without using L'Hopital's rule

Question

I tried:

$$\lim_{x \rightarrow 0} \frac{\ln (x+1)}{\ln (2x+1)} = \frac{1}{\ln(x+1)} \cdot \ln (2x+1) = \ln (2x+1)^{\frac{1}{\ln(x+1)}} = ???$$

What do I do next? I feel like I could use $\lim_{x \rightarrow 0}\frac{\ln (x+1)}{x}=1$ but I am not sure how.

Answer 1

Hint:

$$\frac{\ln(x+1)}{\ln(2x+1)}=\frac{\ln(x+1)}x\frac{2x}{\ln(2x+1)}\frac x{2x}$$