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.
