Finding the limit of $\frac{1}{t\sqrt{1+t}} - \frac{1}{t}$ as $t$ tends to $0$ $$\lim_{t\rightarrow 0}\left(\frac{1}{t\sqrt{1+t}} - \frac{1}{t}\right)$$
I attemped to combine the two fraction and multiply by the conjugate and I ended up with:
$$\frac{t^2-t^2\sqrt{1+t}}{t^3+{t\sqrt{1+t}({t\sqrt1+t})}}$$
I couldn't really work it out in my head on what to do with the last term $t\sqrt{1+t}({t\sqrt{1+t}})$ so I left it like that because I think it works anyways. Everything is mathematically correct up to this point but does not give the answer the book wants yet. What did I do wrong?

Perhaps you were trying something like 
$\dfrac{1}{t\sqrt{1+t}} - \dfrac{1}{t} = \dfrac{1-\sqrt{1+t}}{t\sqrt{1+t}} = \dfrac{1-(1+t)}{t\sqrt{1+t}(1+\sqrt{1+t})} = \dfrac{-1}{\sqrt{1+t}(1+\sqrt{1+t})} $
which has a limit of $\dfrac{-1}{1 \times (1+1)} = -\dfrac{1}{2}$ as $t$ tends to $0$.
Added: If you are unhappy with the first step, try instead $\dfrac{1}{t\sqrt{1+t}} - \dfrac{1}{t} = \dfrac{t-t\sqrt{1+t}}{t^2\sqrt{1+t}} = \dfrac{t^2-t^2(1+t)}{t^3\sqrt{1+t}(1+\sqrt{1+t})} = \dfrac{-t^3}{t^3\sqrt{1+t}(1+\sqrt{1+t})} $ $= \dfrac{-1}{\sqrt{1+t}(1+\sqrt{1+t})}$ to get the same result
