Question #cdd90

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Answer 1 · 2016-06-21T17:59:20

#0#

the cheekiest way is to try and cram it within the scope of L'Hopital's Rule for indeterminate forms, basically #0/0# or #\infty/\infty#

It works as follows:
From Wikipedia

But #lim_(x->0)xlnx = 0 \times -\infty#, no good :-(

We need to play with it a bit. What about #lim_(x->0) (lnx)/(1/x) # which is the same thing. Well that # = - \infty/\infty # .... bingo!!

One application of L'Hopital gives: #lim_(x->0) (lnx)/(1/x) = lim_(x->0) (1/x)/(- 1/x^2) = - lim_(x->0) x = 0#