What is the limits?

There are two ways of doing this.

We will use L'hopital's rule, which states that for #lim_(x->a) f(x)/g(x) = lim_(x->a) (f'(x))/(g'(x))#.

Then...

#d/dx (x ln x) = d/dx (x)*lnx + x *d/dx (lnx) = 1*lnx + x/x = lnx +1#

#d/dx (x^2+1) = 2x...#

#(f'(x))/(g'(x)) = (lnx +1)/(2x)#

This is still not conclusive, so we must perform another iteration of L'hopital's rule...

#d/dx (ln(x)+1) = 1/x, d/dx 2x = 2#

Then #(f''(x))/(g''(x)) = (1/x)/2 = 1/(2x)#

The limit of this function, as easily observed, will be #lim_(x->oo) 1/(2x) = 0#

The second method requires less work, but also requires a stronger understanding. We know that #ln(x) < x# for all x, and that #ln x# increases very slowly by comparison to #x#. We could, depending on how careful we wanted to be, essentially ignore ln x in the numerator for this purpose; the main determining factor in the magnitude of the numerator as #x->oo# is #x#. We can also ignore the +1 on the bottom, as it will rapidly become negligible. Then, if we ignore #lnx#...

#lim_(x->oo) (xlnx)/(x^2+1) approx lim_(x->oo)(x)/(x^2) = lim_(x->oo) 1/x = 0#