Computing $\lim_{n\to\infty}\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )^n$ I am struggling to find the limit of $(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1} )^n$ as $n$ goes to $\infty$.
I know that the limit of $(1+\frac{t}{n})^n$ as $n$ goes to $\infty$ equals $e^t$ and that my limit should approach $e^{t-1}$. 
However, I am having a hard time using and manipulating the known $(1+\frac{t}{n})^n$ limit to find the value that my limit approaches.
I tried $\lim_{n \to \infty} (\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1} )^n=(\lim_{n \to \infty} (\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\lim_{n \to \infty} \frac{n}{n+1})^n$ to avoid using $\lim_{n \to \infty}(1+\frac{t}{n})^n=e^t$ but I am unsure if it is correct.
If possible, I would prefer a very simple method that just uses properties of limits.
Any help is appreciated, thank you in advanced.

In the same spirit as other answers, you could even go beyond the limit.
Starting with $$A=\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )^n$$
$$\log(A)=n \log\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )$$ Now, using long division (or better, Taylor series)
$$\frac{n}{(n+1)^2(n+2)}=\frac 1 {n^2}-\frac{4}{n^3}+O\left(\frac{1}{n^4}\right)$$
$$\sqrt{\frac{n}{(n+1)^2(n+2)}}=\frac 1 {n}-\frac{2}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$\frac{n}{n+1}=1-\frac 1 {n}+\frac{1}{n^2}+O\left(\frac{1}{n^3}\right)$$ making
$$\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}=1+\frac{t-1}{n}+\frac{1-2 t}{n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )=\frac{t-1}{n}+\frac{-\frac{t^2}{2}-t+\frac{1}{2}}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(A)=(t-1)+\frac{-\frac{t^2}{2}-t+\frac{1}{2}}{n}+O\left(\frac{1}{n^2}\right)$$ $$A=e^{\log(A)}=e^{t-1}\left(1-\frac{t^2+2t-1 }{2n}\right)+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
