My question concerns a method that was used to prove the following statement $$ \lim_{A\to\infty} \frac{1}{A}\int_1^A A^{\frac{1}{x}} \ dx = 1. $$ 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 $A > A_0$, $1 + \delta < K\log A < A$ 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 $[1, 1+\delta], [1+\delta, K\log A], [K\log A, A]$ and estimating the integrand on each piece. First sending $A\to\infty$ and then Sending $\delta \to 0$ and $K \to \infty$, we obtain $$ 1 \le \liminf_{A\to \infty} \frac{1}{A}\int_1^A A^{\frac{1}{x}} \ dx \le \limsup_{A \to \infty} \frac{1}{A}\int_1^A A^{\frac{1}{x}} \ dx \le 1. $$
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.
Here comes a long story, the question addresses the choice of a special shape (for some proof of a relation), this is not a mathematical issue, is rather similar to the choice of the narrative art in an essay. And for the answer, i feel that a story is the best way to understand the situation.
We have to show a relation. The relation involves an integral. Under the integral we have a function. This particular function has its kind of behaviour near one, "near infinity", and in between. Often, for didactical reasons, and/or in some analysis courses at the very beginning of studying integration, and/or as an exercise for estimations an explicit computation of an expression involving an integral is not wanted or explicitly made difficult. A problem wants also only some estimative or limit information on the integral. In such cases, one has to (generously) estimate, and this process may break into cases. This seems to break the estetic part, but does not break the pragmatism. Finally, one can still use compact estimations to recover estetics.
In such cases, i always want to explicitly estimate some explicit integral, i want to see the number. Let's do this as a first step for $$ J = \frac 1{2021}\int_1^{2021} 2021^{1/x}\; dx\ . $$ The estimation $1 = \frac 1{2021} \int_1^{2021} 1^{1/x}\; dx\le J$ is clear, in general we have the same simple argument on this side. So let's work on the other side. We want to involve only simple estimations. The function $2021^{1/x}$ has to be estimated "easily" (in the first hours of an analysis course), so that there is not much to compute after. My first rough try is maybe to compare with (either) $2021^{1/1}$, $2021^{1/2}$, $2021^{1/3}$, $2021^{1/4}$, ... and so on. Of course, a small piece at the start of the interval has to be broken into a special case / term, and using for instance $2021^{1/33}$ we would get $$ J \le \frac 1{2021}\left(\int_1^{33} 2021^{1/x}\; dx+\int_{33}^{2021} 2021^{1/x}\; dx\right) \le \frac{33}{2021}\cdot 2021 + 2021^{1/33}\ . $$ $33$ is maybe not optimal, but the idea of estimation may now become clear, it does not work, but we would like to make this work, and do in general "the same". The first term is "bad", it is that big $33$, we were too generous while using $1/x\le 1/1$, and doing "this" in general does not work. Maybe we should slightly refine and try instead the better splitting... $$ \begin{aligned} J &\le \frac 1{2021} \left( \int_1^{2} 2021^{1/x}\; dx + \int_2^{33} 2021^{1/x}\; dx + \int_{33}^{2021} 2021^{1/x}\; dx\right) \\ &\le \frac 1{2021} \left( \int_1^{2} 2021^{1/1}\; dx + \int_1^{21} 2021^{1/2}\; dx + \int_1^{2021} 2021^{1/21}\; dx\right) \\ &\le \frac{(2-1)}{2021}\cdot 2021 + \frac{33}{2021}\cdot 2021^{1/2} + 2021^{1/33}\ . \end{aligned} $$ This may work in general.
We have to arrange that the first term is small, so the first integral is taken on a small interval starting in $1$. OK, let us take some $\delta$ and use $[1,1+\delta]$ for the first interval. The corresponding integral (divided by $2021)$ is estimated above by delta.
Now we have to arrange that the second estimative term above is small. This term is in the special $33$-case above $$\displaystyle \frac 1{2021}\int_1^{33} 2021^{1/(1+\delta)} \le 33\cdot 2021^{\frac 1{1+\delta}-1} = 33\cdot 2021^{\frac \delta{1+\delta}} \le 33\cdot 2021^{-\delta} = \frac{33}{2021^\delta}\ . $$ Can we make this part in a general argument small? So that the following third term is also under estimated control? Yes. Let's see how. Since psychologically a big A is not big enough for my eyes, i will use an $N$, related by $$ A = N^{110}\ . $$ I did not want to write $A=N^K$, and have one more letter $K$, for the story it may seem better to have a number as exponent. Later, there will be an issue to find the "right ballance" between $N$ and $K$. Now we split and estimate almost by only copy+paste+change: $$ \begin{aligned} J &\le \frac 1{N^{110}} \left( \int_1^{1+\frac 1{10}} (N^{110})^{1/x}\; dx + \int_{1+\frac 1{10}}^N (N^{110})^{1/x}\; dx + \int_N^{N^{110}} (N^{110})^{1/x}\; dx\right) \\ &\le \frac 1{N^{110}} \left( \int_1^{1+\frac 1{10}} (N^{110})^{1/1}\; dx + \int_{1+\frac 1{10}}^N (N^{110})^{1\Big/\left(1+\frac 1{10}\right)}\; dx + \int_N^{N^{110}} (N^{110})^{1/N}\; dx\right) \\ &\le \frac 1{10} + \frac 1{N^{110}}\cdot N\cdot (N^{110})^{\frac {10}{11}} + \underbrace{\frac 1{N^{110}}\cdot (N^{110}-N)}_{\le1}\cdot N^{\frac {110}N} \\ &\le \frac 1{10} + \frac 1{N^{10-1}} + N^{\frac {110}N} \ . \end{aligned} $$ The first two terms are easily arranged in the general estimation. For the third term, we come need to find the "right ballance" between $N$ and $K$, when dealing with an Ansatz like $A=N^K$, with independent choices $N\to\infty$, $K\to\infty$. (The very special case $K=110$ was considered above, but a similar estimation applies in general.) For instance, if $K\sim\sqrt N$ we are done, because $N^{1/\sqrt N}\to 1$, because $\sqrt N^{1/\sqrt N}\to 1$.
This story was long, worse, the same idea repeats itself, but i think such an answer has to be written somewhere to show how much work is behind a solution inside the $\epsilon-\delta$-avatar, and why it takes some time to learn when left on own devices. We stress for the purposes of the OP the features, that now make the answer:
We have to estimate an integral. Without computing. Reason: beginning of an analysis course, for instance. Also, the involved function $A^{1/x}$ under the integral should be not rewritten as $\exp(\log A/n)$, followed by calculus estimation, same reason, we are at the start of a course. Simple estimations are wanted. E.g. from $x\ge 1$ we estimate $A^{1/x}\le A^{1/1}=A$. This does not work, but we use this plain idea separately on some more intervals, the estimation is thus done with a step function with conveniently chosen steps. The result of the estimation is finally also a simple expression like $\sqrt[n]n$, and it is easy to show it converges to one at this beginning of the course.
In this sense, i would answer the given explicit questions as follows:
(1) What motivates the division of the region of integration into 3 pieces?
The solution had some very special estimation strategy, if we know this start, we finally land with the three intervals. But the strategy to estimate is the essence.
(2) Is there an intuitive explanation as to why separate bounds on the different regions are effective?
Yes, the long story above tries to explain how we arrange in an interactive thinking process the estimation. The same and again the same idea is refined and made to work. We have a puzzle with our parameters in the estimation, if we manage to get them in the right order, the solution works. In practice, the function to be estimated under the integral determines the situation. (And of course, our skills.)
Books and courses have no space and no time to show the thinking process, even worse, after having a solution that works and fits the end of the thinking process, it is further "compactified" to fit in time + space. Often important didactical steps get lost.
(3) 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.
No, i would say. Not in essence. Often, estimations need simple and brutal = generous steps. Such steps do not work for all points, but combined ideas finally work, so the solution uses one estimation on some region, an other one on an other region.