Calculate $\lim_{x \to \infty}{\frac{\log(2x+1)}{\log(3x+2)}}$

Question

Calculate $\lim_{x \to \infty}{\dfrac{\log(2x+1)}{\log(3x+2)}}$

I've used L'Hôpital's rule and the solution is $1$. However, can I calculate it without using L'Hôpital's rule?

Answer 1

You could try this:

$$\log (2x+1)= \log 2x+\log \left(1+\frac 1{2x}\right)=\log x+\log 2+\log \left(1+\frac 1{2x}\right)$$ and similarly for the denominator - ie isolating the part which grows fastest.

Answer 2

You don't need use L'Hôpital's rule if you are allowed to use the fact that $\log(kx+1)\approx \log(kx)$ as $x\to +\infty$ for $k>0$, then re-write $$\frac{\log(ax+1)}{\log(bx+1)}\approx \frac{\log(ax)}{\log(bx)}=\frac{\log(a)+\log(x)}{\log(b)+\log(x)}=\frac{\frac{\log(a)}{\log(x)}+1}{\frac{\log(b)}{\log(x)}+1}$$ for $a,b>0$ then setting $x\to +\infty$ and the result is follows.