What is $ \lim_{n \to \infty} n^{(n+1)/n} (n+1)^{(-n-2)/(n+1)} $

Question

How to evaluate this limit? Is it okay to evaluate first the limit of the exponents, and then the limits of the bases?

$$ \lim_{n \to \infty} n^{(n+1)/n} (n+1)^{(-n-2)/(n+1)} $$

Answer 1

From above Hint: $$ \lim_{n\to\infty}n^{(n+1) / n}(n+1)^{(-n-2) /(n+1)} = \lim_{n\to\infty} \frac{n}{n+1}\frac{n^{\frac 1n}}{(n+1)^{\frac{1}{n+1}}} $$ Use $\lim_{n\to\infty}\frac{\ln n}{n} = 0$ , so $\lim_{n\to\infty}\ln n^{\frac 1n} = 0$ , and then $\lim_{n\to \infty} n^{\frac 1n} = 1$

Answer 2

Another "common" hint:

$$\lim_{n \to \infty} n^{(n+1)/n} (n+1)^{(-n-2)/(n+1)} = \lim_{n \to \infty} e^{\log \left[n^{(n+1)/n} (n+1)^{-(n+2)/(n+1)}\right]}$$

You have reduced the limit to this one and using log rules you need to find

$$ \lim_{n \to \infty} \left[\left(\frac{n+1}{n}\right)\log(n) - \frac{n+2}{n+1}\log(n+1)\right] \tag{$\star$}$$

Now you can make a common denominator and with some simple asymptotics you get that the limit in $\star$ is equal to $0$, thus the original one is $e^0 = 1$