We know that for large values of x, higher powers always eventually prevail over lower powers. Thus we can claim that if x is sufficiently large, then x²/2 ≥ 2x. This can be confirmed by finding the set on which the desired inequality is true and see that it includes the area we need; in our case it is easy, in fact the inequality is true for all x ≥ 4.

Sometimes the precise solution would be difficult; then we can for instance use the notion of limit. In our example we would show that

which in particular means that there is a constant K (we may assume that it is larger than 3) such that for all x ≥ K, the ratio is equal to at least one. This means that for x ≥ K we indeed have x²/2 ≥ 2x.

Whatever justification you use, for x ≥ K we have

This imples that also

Since the integral on the right converges, by the Comparison test, the integral of the given function from K to infinity is also convergent. Between 3 and K there are no problems, so the given function is Riemann integrable there. Putting these two facts together we see that the given integral (from 3 to infinity) is convergent.

Now that we know that the given integral converges, we can use the failed attempt at the Comparison test to obtain a lower estimate for its value: It is at least 1/3.