Intuitively, we should know that the given sequence converges to #0# as any form of polynomial growth (quadratic in this case) is significantly faster than logarithmic growth. This should make sense as being the inverse of an exponential function, a logarithmic function must increase very slowly.
However, while this gives us an idea of what the result should be, it is not a proof. Let's show the result more formally. To do so, we will look at a function with the same end behavior as #a_n#.
Let #f(x)= ln(x)/(x^2 + 1)#.
By the construction of #f(x)#, it should be clear that if #f(x)# converges as #x->oo#, then #a_n# converges as #n->oo#, and #lim_(n->oo)a_n = lim_(x->oo)f(x)#.
(Note that this can also be proven formally with an #epsilon-delta# proof, but such detail is likely unnecessary for a student of calculus).
Now, all that remains is to show that #lim_(x->oo)f(x) = 0#.
As #lim_(x->oo)ln(x) = lim_(x->oo)(x^2+1) = oo#, we may apply L'hospital's rule to #f(x)# as an #oo/oo# case.
#lim_(x->oo)ln(x)/(x^2+1) = lim_(x->oo)(d/dxln(x))/(d/dx(x^2+1))#
#=lim_(x->oo)(1/x)/(2x)#
#=lim_(x->oo)1/(2x^2)#
#= 1/oo#
#=0#
And so
#lim_(n->oo)a_n = lim_(x->oo)f(x) = 0#
