Originally Posted by
ThePerfectHacker Still no good you should have mentioned the open interval $\displaystyle (-1,1)$
Anyway, the way you show continuity is to show that,
$\displaystyle \lim_{x\to 0}f(x)=f(0)=0$
Since,
$\displaystyle f(x)=\frac{1}{\ln|x|}$ open some open interval except possibly at $\displaystyle x=0$ we have,
$\displaystyle \lim_{x\to 0}\frac{1}{\ln|x|}=0$
Which is true if we consider the left-right limits,
$\displaystyle \lim_{x\to 0^-}\frac{1}{\ln (-x)}=\lim_{x\to 0^+}\frac{1}{\ln x}$
And by using L'Hopital's Rule.