How do you find the limit of # xln(x)# as x approaches 0?

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Answer 1 · 2016-12-18T17:31:39

I have not found an algebraic solution.

Using l'Hospital's rule, we need to rewrite first to get indeterminate form #0/0# or #+- oo/oo#

#lim_(xrarr0)xlnx# has initial form #0 (-oo)#

Rewrite as #lim_(xrarr0)lnx/(1/x)#

Now apply l'Hospital

#lim_(xrarr0)lnx/(1/x) = lim_(xrarr0)(1/x)/(-1/x^2)# provided the second limit exists or is #+-oo#.

#lim_(xrarr0)(1/x)/(-1/x^2) = lim_(xrarr0)(-x) = 0#.

So, #lim_(xrarr0)xlnx = lim_(xrarr0)lnx/(1/x) =0#