# real analysis – Bounding an integral by splitting the interval of integration

My question concerns a method that was used to prove the following statement
$$lim_{Atoinfty} frac{1}{A}int_1^A A^{frac{1}{x}} dx = 1.$$
The method is as follows. First we obtain the easy lower bound
$$frac{1}{A}int_1^A A^{frac{1}{x}} dx > 1 – frac{1}{A}, A > 1$$
and then to get a tight upper bound, we show that for every $$delta > 0$$ and for every $$K > 0$$ there exists $$A_0(delta, K) > 1$$ such that for all $$A > A_0$$, $$1 + delta < Klog A < A$$ and
$$frac{1}{A}int_1^A A^{frac{1}{x}} dx < delta + A^{-frac{delta}{1+delta}}log A + e^{frac{1}{K}}.$$

We prove the last statement by dividing the interval of integration into $$3$$ pieces $$(1, 1+delta), (1+delta, Klog A), (Klog A, A)$$ and estimating the integrand on each piece. First sending $$Atoinfty$$ and then Sending $$delta to 0$$ and $$K to infty$$, we obtain
$$1 le liminf frac{1}{A}int_1^A A^{frac{1}{x}} dx le limsup frac{1}{A}int_1^A A^{frac{1}{x}} dx le 1.$$

My questions are: what motivates the division of the region of integration into $$3$$ pieces? Is there an intuitive explanation as to why separate bounds on the different regions are effective? Is this a general method that works in other situations as well? If so, I would love to see a sketch of an example and/or a general principle on splitting up regions of integration to obtain desired bounds.