I tried to solve this limit: $\lim\limits_{n \to +\infty} \bigg(\dfrac{n+1}{n+2}\bigg)^n$.
My approach was to re-write it as $\lim\limits_{n \to +\infty} \bigg(\dfrac{n}{n+2} + \dfrac{1}{n+2}\bigg)^n$, and since $\dfrac{n}{n+2}$ tends to 1 and $\dfrac{1}{n+2} \sim \dfrac{1}{n}$ as $n \to +\infty$, I figured the solution would be $e$, as $\lim\limits_{n \to +\infty} \bigg(1+\dfrac{1}{n}\bigg)^n = e$.
I suppose I've done something wrong, since by plotting the function I noticed the solution is $\dfrac{1}{e}$.
Where is my error?
Best Answer
The error lies in assuming that, since $\lim_{n\to\infty}\frac n{n+2}=1$ and since $\frac1{n+2}\sim\frac1n$, then $\lim_{n\to\infty}\left(\frac n{n+2}+\frac1{n+2}\right)^n=\lim_{n\to\infty}\left(1+\frac1n\right)^n$. To see why it doesn't work, consider the limit$$\lim_{n\to\infty}n\left(\frac n{n+1}-1\right).$$It is equal to $1$, right?! However, by your argument, since $\lim_{n\to\infty}\frac{n+1}n=1$, it should be equal to$$\lim_{n\to\infty}n(1-1)=0.$$