Limit without l'hopital $\lim_{x \to 0} \left ( \frac{1}{x} -\frac{1}{e^x-1}\right )$ I want to find the limit without L'hopital's rule and without  Taylor series:
$
\lim_{x \to 0} \left ( \frac{1}{x} -\frac{1}{e^x-1}\right )
$
Is it possible? 
Hints are more than welcome! 
Thanks!
 A: Note that $\frac{1}{x} - \frac{1}{e^x-1} = \frac{e^x-1-x}{(x)(e^x-1)}$
This, in turn, can be written as: $\frac{e^x-1-x}{x^2} \times \frac{x}{e^x-1}$
I do believe you need to know that $\frac{x}{e^x-1}$ tends to $1$ and $\frac{e^x-1-x}{x^2}$ tends to $\frac{1}{2}$. 
A: Let us rewrite the expression in the limit as follows 
$$ \frac{1}{x} - \frac{1}{e^x-1} = \frac{e^x - 1 - x}{xe^x - x} $$
Using the Taylor expansion of $e^x$, which we recall is 
$$ e^x = \sum_{i = 0}^{\infty} \frac{x^i}{i!} \ , $$ 
we further rewrite the numerator and denominator as
$$ e^x - 1 - x = \frac{x^2}{2} + \mathcal{O} (x^3) = x^2(\frac{1}{2} +  \mathcal{O} (x) )\\ xe^x - x = x^2 + \mathcal{O} (x^3) =  x^2( 1 + \mathcal{O}(x) ) $$
where $ \mathcal{O} (x^3) $ denotes the terms of order higher then $2$ in the variable $x$. Since $\lim_{x \rightarrow 0} \mathcal{O}(x) = 0$ we can compute, using the theorem for product of limits, in the following manner
$$ \lim_{x \rightarrow 0} (\frac{1}{x} - \frac{1}{e^x-1} ) = \lim_{x \rightarrow 0} (\frac{e^x - 1 - x}{xe^x - x} ) = \lim_{x \rightarrow 0} \frac{x^2(\frac{1}{2} +  \mathcal{O} (x) )}{x^2(1 +  \mathcal{O} (x) )} = \lim_{x \rightarrow 0} \frac{\frac{1}{2} +  \mathcal{O} (x)}{1 +  \mathcal{O} (x)} = \frac{1}{2} $$
A: Note that$$\lim_{x\to0}\frac1x-\frac1{e^x-1}=\lim_{x\to0}\frac{1-\frac x{e^x-1}}x.$$Now, since$$\frac{e^x-1}x=1+\frac x2+O(x^2),$$you have$$\frac x{e^x-1}=1-\frac x2+O(x^2)$$and therefore$$1-\frac x{e^x-1}=\frac x2+O(x^2).$$Therefore, your limit is $\frac12$.
