<s>[INST] 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.
 [/INST] 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.
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