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?
 A: 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
A: Asymptotics:  
$$\begin{align}
\frac{1}{\sqrt{1+t}} &= (1+t)^{-1/2} = 1 - \frac{1}{2}\;t + o(t)
\\
\frac{1}{t\sqrt{1+t}} &= \frac{1}{t} - \frac{1}{2} + o(1)
\\
\frac{1}{t\sqrt{1+t}} - \frac{1}{t} &= - \frac{1}{2} + o(1) .
\end{align}$$  
A: I'd use a substitution to get rid of the surd.
$$\mathop {\lim }\limits_{t \to 0} -\frac{1}{t}\left( {1 - \frac{1}{{\sqrt {t + 1} }}} \right) = $$
$$\sqrt {t + 1}  = u$$
$$\mathop {\lim }\limits_{u \to 1} -\frac{1}{{{u^2} - 1}}\left( {1 - \frac{1}{u}} \right) = $$
$$\mathop {\lim }\limits_{u \to 1} -\frac{1}{{{u^2} - 1}}\left( {\frac{{u - 1}}{u}} \right) = $$
$$\mathop {\lim }\limits_{u \to 1} -\frac{1}{{u + 1}}\left( {\frac{1}{u}} \right) = -\frac{1}{2}$$
A: You could also use L'Hopitals rule:
First note that
$\frac{1}{t\sqrt{1+t}} - \frac{1}{t} = \frac{1-\sqrt{1+t}}{t\sqrt{1+t}}$
L'Hopitals rule is that if: $f(x)=0$ and $g(x)=0$ then
$\lim_{t\to x} \frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)}$
with some provisos that I'll ignore here...
In our case


*

*$f(t) = 1 - \sqrt{1+t}$
So $f'(t) = (-1/2)(1+t)^{-1/2}$ and $f'(0)=-1/2$.

*$g(t) = t\sqrt{1+t}$ 
So $g'(t) = \sqrt{1+t} + (t/2)(1+t)^{-1/2}$ and $g'(0)=1$
So finally we get $f'(0)/g'(0) = -1/2$ as the limit we need.
A: Let $f:]0,\infty[\to\mathbb{R}$ given by $$f(x)=\frac{1}{\sqrt{x}}.$$ Then $$\frac{1}{t\sqrt{1+t}} - \frac{1}{t}=\frac{f(1+t)-f(1)}{t},$$
so $$\lim_{t\to 0} \frac{1}{t\sqrt{1+t}} - \frac{1}{t}=\lim_{t\to 0} \frac{f(1+t)-f(1)}{t}=f'(1).$$
Since $$f'(x)=-\dfrac{1}{2}\cdot x^{-\frac{3}{2}}$$ in $]0,\infty[,$ we get
$$\lim_{t\to 0} \frac{1}{t\sqrt{1+t}} - \frac{1}{t}=\left. -\dfrac{1}{2}\cdot t^{-\frac{3}{2}}\right|_1=-\frac{1}{2}.$$
