$\lim_{n\rightarrow\infty}(a+\frac{n+1}{bn^2+n+2})^{n+2}=\frac1e$
So I tried to get to the common denominator and then add 1 and substract 1 inside the power to create the case $1^\infty$. But I get to something really absurd and wrong. What's wrong with my approach?
On
The right approach is to take logs here. By the given conditions we have $$\lim_{n\to\infty} (n+2)\log\left (a+\frac{n+1}{bn^2+n+2}\right) =-1\tag{1}$$ and this implies that the log term in above equation tends to $0$. Therefore the argument of the log term tends to $1$ ie $$a+\frac{n+1}{bn^2+n+2}\to 1\tag{2}$$ If $b\neq 0$ then this means that $a=1$ and then from equation $(1)$ we get $$\lim_{n\to\infty} \frac{(n+2)(n+1)}{bn^2+n+2}\cdot \frac{bn^2+n+2}{n+1}\log\left(1+\frac{n+1}{bn^2+n+2}\right)=-1$$ ie $1/b=-1$ so that $b=-1$.
All of the above is based on the assumption that $b\neq 0$. Let's see what happens when $b=0$. Clearly in that case argument of logarithm tends to $a+1$ so that $a=0$. It is easy to see that in this case also the equation $(1)$ holds so that we have two sets of values for $a, b$ which satisfy the given conditions: $a=1,b=-1$ and $a=b=0$.
Recall that the limit for $e^x$ is as follows:
$$\lim_{n \to \infty} \left(1+{x \over n}\right)^{n}$$
So to achieve $\frac{1}{e}$, $x$ must be $-1$. You get this by using $a=1$ and $b=-1$ and factoring:
$$\require{cancel} \lim_{n\rightarrow\infty}\left(1+\frac{n+1}{-n^2+n+2}\right)^{n+2}=\lim_{n\rightarrow\infty}\left(1+\frac{\cancel{n+1}}{-1(n-2)(\cancel{n+1})}\right)^{n+2}=\lim_{n\rightarrow\infty}\left(1-\frac{1}{n-2}\right)^{n+2}$$
And using some manipulation:
$$\lim_{n\rightarrow\infty}\left(1-\frac{1}{n-2}\right)^{n+2}=\lim_{n\rightarrow\infty}\left[\left(1-\frac{1}{n-2}\right)^{n-2}\right]^{n+2\over n-2}= \left({1 \over e}\right)^1$$