Evaluate $\lim_{t\rightarrow+\infty}{\left(1+\frac{1}{2+t}\right)\left({\frac{t^2-2t-1}{-t-2}}\right)+t}$ I want to evaluate the following limit
$$\lim_{t\rightarrow+\infty}{\left(1+\frac{1}{2+t}\right)\left({\frac{t^2-2t-1}{-t-2}}\right)+t}$$
I know the result is $3$. It's very tempting for me to do $\frac{t^2-2t-1}{-t-2}\sim-t$ but I know this is not allowed because of the final $+t$ there.
I've tried multiplying the brackets like this:
$$\lim_{t\rightarrow+\infty}{\left(1+\frac{1}{2+t}\right)\left({\frac{t^2-2t-1}{-t-2}}\right)+t}=\lim_{t\rightarrow+\infty}{\frac{t^2-2t-1}{-t-2}}+\frac{t^2-2t-1}{(2+t)(-t-2)}+t=\dots
$$
Any hints on how to find the limit?
 A: Hint: Use that your term is equal to $${\frac {3\,{t}^{2}+11\,t+3}{ \left( 2+t \right) ^{2}}}$$ and this is equal to
$$\frac{3+\frac{11}{t}+\frac{3}{t^2}}{1+\frac{4}{t}+\frac{4}{t^2}}$$
A: We have $$\left(1+\frac{1}{2+t}\right)\left({\frac{t^2-2t-1}{-t-2}}\right)+t=-\frac{(t+3)(t^2-2t-1)}{(t+2)^2}+\frac{t(t+2)^2}{(t+2)^2}$$ or $$-\frac1{(t+2)^2}[t^3+t^2-7t-3-t^3-4t^2-4t]=\frac{3t^2+11t+3}{t^2+4t+4}$$ which can be written as $$\frac{3+\frac{11}t+\frac3{t^2}}{1+\frac4t+\frac4{t^2}}\to\frac{3+11\cdot0+3\cdot0}{1+4\cdot0+4\cdot0}=3$$ as $t\to+\infty$.
A: There are a few ways you can tackle this. Using the working you've done,
\begin{align*}
&\frac{t^2-2t-1}{-t-2} + t + \frac{t^2-2t-1}{(2+t)(-t-2)} \\
= \; &\frac{t^2-2t-1}{-t-2} + \frac{t(-t - 2)}{-t - 2} + \frac{t^2-2t-1}{(2+t)(-t-2)} \\
= \; &\frac{t^2-2t-1}{-t-2} + \frac{-t^2 - 2t}{-t - 2} + \frac{t^2-2t-1}{(2+t)(-t-2)} \\
= \; &\frac{-4t-1}{-t-2} + \frac{t^2-2t-1}{(2+t)(-t-2)} \\
= \; &\frac{-4-\frac{1}{t}}{-1-\frac{2}{t}} + \frac{1-\frac{2}{t}-\frac{1}{t^2}}{(\frac{2}{t}+1)(-1-\frac{2}{t})} \\
\to \; & 4 - 1 = 3
\end{align*}
