How to calculate the limit $\lim_{x\to\infty}\frac{\ln(2+x)}{\ln(3+x)}$ without L'Hôpital nor a series expansion?

Question

Find the following limit without L'Hôpital nor a series expansion $$\lim_{x\to\infty}\frac{\ln(2+x)}{\ln(3+x)}$$

I tried to use the substitution $x=\frac{1}{y}$, under which the limit is $$\lim_{y\to 0^+}\frac{\ln(1+2y)-\ln(y)}{\ln(1+3y)-\ln(y)}$$ but it did not help to solve it.

Answer 1

$$\lim_{x\rightarrow+\infty}\frac{\ln(2+x)}{\ln(3+x)}=\lim_{x\rightarrow+\infty}\frac{\ln{x}+\ln\left(\frac{2}{x}+1\right)}{\ln{x}+\ln\left(\frac{3}{x}+1\right)}=$$ $$=\lim_{x\rightarrow+\infty}\frac{1+\frac{\ln\left(\frac{2}{x}+1\right)}{\ln{x}}}{1+\frac{\ln\left(\frac{3}{x}+1\right)}{\ln{x}}}=\frac{1+\lim\limits_{x\rightarrow+\infty}\frac{\ln\left(\frac{2}{x}+1\right)}{\ln{x}}}{1+\lim\limits_{x\rightarrow+\infty}\frac{\ln\left(\frac{3}{x}+1\right)}{\ln{x}}}=\frac{1+0}{1+0}=1.$$

Answer 2

Hint: $$ \begin{align} \lim_{x\to\infty}\frac{\log(2+x)}{\log(3+x)}-1 &=\lim_{x\to\infty}\frac{\log\left(\frac{2+x}{3+x}\right)}{\log(3+x)}\\ \end{align} $$

Answer 3

Let $3+x=e^y$ with $y\to+\infty$

$$\frac{\log(2+x)}{\log(3+x)}=\frac{\log(e^y-1)}{\log e^y}=\frac{\log e^y+\log(1-\frac1{e^y})}{\log e^y}=1+\frac{\log(1-\frac1{e^y})}{y}\to1+\frac{0}{+\infty}=1$$