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What is the limit of #lnx# as x approaches #0#?

1 Answer
Oct 22, 2016

Answer:

#lim_(xrarr0)lnx=-oo#, ie the limit does not exists as it diverges to #-oo#

Explanation:

You may not be familiar with the characteristics of #ln x# but you should be familiar with the characteristics of the inverse function, the exponential #e^x#:

Let # y=lnx=> x = e^y #, so as # xrarr0 => e^yrarr0#

You should be aware that #e^y>0 AA y in RR#,but #e^yrarr0# as #xrarr-oo#.

The graph of #f(x)=e^x# should help illustrate this:
graph{e^x [-10, 10, -5, 5]}

so if we want #e^yrarr0=>yrarr-oo#

Therefore we can conclude that #lim_(xrarr0)lnx=-oo#, ie the limit does not exist as diverges to #-oo#

The graph of #f(x)=lnx# should help illustrate this:
graph{lnx [-10, 10, -5, 5]}