Let f(x)=((ln|x|)/x)^(1/x) (I'm assuming you meant to use the natural logarithm here, though the approach used can be easily modified to use the common logarithm).
Then ln(f(x))=1/x * ln((ln|x|)/x)=(ln(ln|x|)-ln(x))/x, by properties of logarithms.
The limit lim_{x->infty}ln(f(x)) is thus an infty/infty indeterminate form , to which L'Hopital's Rule can be applied.
Since d/dx(ln(ln|x|)-ln(x))=1/(ln|x|)*1/x-1/x and d/dx(x)=1, we can say that
lim_{x->infty}ln(f(x))=lim_{x->infty}(1/(ln|x|)*1/x-1/x)/1=0/1=0.
The continuity of the natural log function can now be used to say that this means ln(lim_{x->infty}f(x))=0 so that lim_{x->infty}f(x)=e^{0}=1.