#f(x)=ln(x)-> \infty# as #x->\infty# (#ln(x)# grows without bound as #x# grows without bound) and #f(x)=ln(x)->-infty# as #x -> 0^{+}# (#ln(x)# grows without bound in the negative direction as #x# approaches zero from the right).

To prove the first fact, you essentially need to show that the increasing function #f(x)=\ln(x)# has no horizontal asymptote as #x->\infty#.

Let #M>0# be any given positive number (no matter how big). If #x>e^{M}#, then #f(x)=ln(x)>ln(e^{M})=M# (since #f(x)=ln(x)# is an increasing function). This proves that any horizontal line #y=M# cannot be a horizontal asymptote of #f(x)=\ln(x)# as #x->\infty#. The fact that #f(x)=\ln(x)# is an increasing function now implies that #f(x)=\ln(x)->\infty# as #x->infty#.

To prove the second fact, let #M>0# be any given positive number so that #-M<0# is any given negative number (no matter how far from zero). If #0 < x < e^{-M}#, then #f(x)=ln(x)<\ln(e^{-M})=-M# (since #f(x)=ln(x)# is increasing). This proves that #f(x)=\ln(x)# gets below any horizontal line if #0 < x# is sufficiently close to zero. That means #f(x)=ln(x)->-infty# as #x -> 0^{+}# .