Originally Posted by
Bacterius $\displaystyle f(x) = \frac{ 1}{(x+1)^{1/x}} - \frac{ 1}{x}$
You want :
$\displaystyle \lim_{x\to 0} f(x)$
Therefore, you want :
$\displaystyle \lim_{x\to 0} f(x) = \frac{ 1}{(x+1)^{1/x}} - \frac{ 1}{x}$
Let's not forget that $\displaystyle \lim_{x\to 0}\frac{ 1}{x} = \pm \infty$ (depending on which side you come from)
Consider if you are approaching $\displaystyle 0$ on the positive side :
$\displaystyle \lim_{x\to 0} f(x) = \frac{ 1}{1^{(+\infty)}} - (+\infty)$
Ugly, eh ?
$\displaystyle \lim_{x\to 0} f(x) = 1 - \infty$
Which by absorption (I'm not sure about the last part) :
$\displaystyle \lim_{x\to 0} f(x) = - \infty$
The other possibility (approaching $\displaystyle 0$ on the negative side) seems quite harder (since you get some weird results) but since I'm not really sure about what I've just done, I won't go on.