Show function is order small o(t) for $t \to 0$ I am trying to solve the following exercise:

Show that$$\frac{1}{e^t-1} = \frac{1}{t} - \frac{1}{2} + \frac{1}{12}t + o(t)$$
  for $t\to 0$.  

So what I need to show is that
$$\frac{1}{e^t - 1} - \frac{1}{t} - \frac{1}{12}t + \frac{1}{2} = o(t)$$
for $t\to 0$ which can be restated as 
$$\lim_{t \to 0}\left |\frac{\frac{1}{e^t - 1} - \frac{1}{t} - \frac{1}{12}t + \frac{1}{2}}{t} \right | = 0$$
which means that I have to show that the function $$f(t) := \frac{1}{t(e^t - 1)} - \frac{1}{t^2} - \frac{1}{12} + \frac{1}{2t}$$ is continuous in $t_0 = 0$ with $f(0) = 0$. However I don't really know what to do from here, I can't find good estimations to work with $\epsilon - \delta$. I am looking for a very basic approach, possibly without methods from differential calculus. I know that $exp(x)$ grows faster than every polynomial and I am pretty sure this is of good use here, but I don't know how.
 A: Note that we can write
$$\begin{align}
\frac{1}{1-e^t}&=\frac{1}{-t\left(1+\frac12t+\frac16t^2+O(t^3)\right)}\\\\
&=\frac{1-\frac12t+\frac{1}{12}t^2+O(t^3)}{-t}\\\\
&=-\frac1t+\frac12-\frac1{12}t+O(t^2)
\end{align}$$
Hence, we have
$$\frac{\frac{1}{1-e^2}+\frac1t-\frac12+\frac1{12}t}{t}=O(t)\to 0\,\,\text{as}\,\,t\to 0$$
A: If you're given the series expansion, it is fairly easy to check it.
First, observe that $\frac{1}{e^t-1}$ diverges as $t$ approaches $0$.  However,
$$
\lim_{t\rightarrow 0}\left(\frac{1}{e^t-1}-\frac{1}{t}\right)
$$ 
can be computed easily using l'Hopital's rule.  In particular,
$$
\lim_{t\rightarrow 0}\left(\frac{1}{e^t-1}-\frac{1}{t}\right)=\lim_{t\rightarrow 0}\left(\frac{t-e^t+1}{t(e^t-1)}\right)=\lim_{t\rightarrow 0}\left(\frac{1-e^t}{(e^t-1)+te^t}\right)=\lim_{t\rightarrow 0}\left(\frac{-e^t}{2e^t+te^t}\right)=-\frac{1}{2}.
$$
Now, we've confirmed the first term and discovered the second term.  In fact, we can consider the Taylor series of 
$$
\frac{t-e^t+1}{t(e^t-1)}.
$$
The first two terms are $-\frac{1}{2}$ and $\frac{1}{12}t$.  Then, the remainder is $o(t)$ using Taylor's remainder theorem since the second derivative of the function is bounded near $t=0$.
