How to determine $\lim_{n\rightarrow \infty} \frac{(n+2)^{\log(n+2)}}{(n+1)^{\log(n+1)}}$? As a part of an exercise in which I have to study the convergence of certain power series, I am stuck finding out this limit:
$$\lim_{n\rightarrow \infty} \frac{(n+2)^{\log(n+2)}}{(n+1)^{\log(n+1)}}$$ 
It seems to me that the limit should equal one, since both functions take very similar values and the $+1$ and $+2$ in the numerator and denominator stop being really relevant for higher values of $n$, but I need a way of reaching that conclusion (or the correct one, if my intuition is wrong, which could very well be the case) analytically. I have tried L'Hôpital, but it doesn't seem to simplify the limit. It actualy becomes quite complicated. I have also tried to make the argument that, for higher values of $n$, $n+2 \approx n+1$ and therefore the limit can be expressed as:
$$\lim_{n\rightarrow \infty} n^{\log(\frac{n+2}{n+1})}$$
But I end up with a $\infty^0$-type indetermination, which would require expressing the limit as:
$$\lim_{n\rightarrow \infty} \frac{\log(\frac{n+2}{n+1})}{1/n}$$
In order to have a 0/0 type indetermination where we can apply L'Hôpital. Each option seems worse than the last one. Anyone can suggest a simpler path? I remember my professor didn't take long to solve this in class, so I must be missing something.
 A: We have that


*

*$(n+2)^{\log(n+2)}=e^{\log^2(n+2)}$

*$(n+1)^{\log(n+1)}=e^{\log^2(n+1)}$
therefore
$$\frac{(n+2)^{\log(n+2)}}{(n+1)^{\log(n+1)}}=e^{\log^2(n+2)-\log^2(n+1)}\to1$$
indeed
$$\log^2(n+2)-\log^2(n+1)=(\log(n+2)+\log(n+1))(\log(n+2)-\log(n+1))=$$
$$=\log(n^2+3n+2)\log\left(1+\frac1{n+1}\right)=\frac{\log(n^2+3n+2)}{n+1}\frac{\log\left(1+\frac1{n+1}\right)}{\frac1{n+1}}\to 0$$
since by $t=1+\frac1{n+1}\to 0$


*

*$\frac{\log\left(1+\frac1{n+1}\right)}{\frac1{n+1}}=\frac{\log(1+t)}{t}\to 1$
and by $\sqrt[n]{p(n)}\to 1$


*

*$\frac{\log(n^2+3n+2)}{n+1}=\frac{\frac1n\log(n^2+3n+2)}{1+1/n}=\frac{\log \sqrt[n]{n^2+3n+2}}{1+1/n}\to \frac{\log 1}{1}=0$
A: Idea:
$$
\lim_{n\to\infty}\frac{(n+2)^{\log(n+2)}}{(n+1)^{\log(n+1)}} =
\lim_{n\to\infty}\frac{(n+2)^{\log(n+2)}}{(n+1)^{\log(n+1)}}\frac{(n+2)^{\log(n+1)}}{(n+2)^{\log(n+1)}} =
\lim_{n\to\infty}\left(\frac{n+2}{n+1}\right)^{\log(n+1)}\lim_{n\to\infty}(n+2)^{\log((n+2)/(n+1))}.
$$
Now, take logs of each limit and apply $\log(1 + x)\approx x$ for small $x$ (see https://www.math24.net/infinitesimals/).
A: Make the change: $n=e^x-2$. Then:
$$\frac{(n+2)^{\log(n+2)}}{(n+1)^{\log(n+1)}}\sim
\frac{(e^x)^{\log(e^x)}}{(e^x-1)^{\log(e^x-1)}}\sim\frac{(e^x)^{\log(e^x)}}{(e^x)^{\log(e^x)}}\to 1$$
