What is #lim_(x->0) xlnx - x^2#?
1 Answer
Apr 10, 2015
#y = xlnx - x^2#
If you merely plug in zero, what you get is undefined.
#lim_(x->0) f(x) = # undefined by simply using#x = 0# .
You can try this:
#y = lnx/(1/x) - x^2#
#= lnx/(1/x) - x/(1/x)#
#= (lnx - x)/(1/x)#
Use L'Hopital's Rule to differentiate this (separately for numerator and denominator), because now it has a nice form for it.
#y = (1/x - 1)/(-1/x^2)#
#= (-x^2)(1/x - 1)#
#= -x + x^2#
Now, set them equal to
#x^2 - x = 0#
So, this limit is equal to zero.
graph{xlnx - x^2 [-1.552, 1.866, -1.019, 0.69]}
