Question

Question #cdd90

Answer 1

`0`

Explanation:

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:

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`