Intuitively, as there is no bound to how large we can make #sqrt(x)# by increasing #x#, we expect that the limit as #x->oo# of #sqrt(x)# would be #oo#. Indeed, if there were such a bound, say #x_0#, then we would arrive at a contradiction, as #sqrt(x_0^2+1) > sqrt(x_0^2)= x_0#.
We can, however, approach the problem in a more rigorous manner.
We say that the limit as #x->oo# of a function #f(x)# is #oo# (alternately #f(x)->oo# as #x->oo#), denoted #lim_(x->oo)f(x)=oo#, if, for every integer #N>0# there exists an integer #M>0# such that #x>M# implies #f(x)>N#.
Less formally, that means that for any real value, #f(x)# will be greater than that value for large enough #x#.
Our claim is that #lim_(x->oo)sqrt(x) = oo#. Let's prove it using the above definition.
Take any integer #N>0#, and let #M=N^2#. Then, for any #x>M#, we have
#sqrt(x) >sqrt(M) = sqrt(N^2) = N#
We have shown that for any integer #N>0# there exists an integer #M>0# such that #x>M# implies #sqrt(x) > N#, thereby proving that #lim_(x->oo)sqrt(x) = oo#.
The above method actually can be used to show that #x^k->oo# as #x->oo# for any #k>0#. If we start with an arbitrary #N>0# and let #M=N^(1/k)#, then for #x>M# we have #x^k > M^k = (N^(1/k))^k=N#. As #sqrt(x) = x^(1/2)#, the above is just a special case of this.
