I have a question about how to prove the limit using $\limsup$. On the other day, I was asked the following problem in the exam:
Let $m$ and $n$ be positive integers. Calculate the following limit, where $m$ is fixed. $$ \lim_{n \to \infty} \int_1^e x^m e^x (\log x)^n dx $$
To this problem, I answered in the manner shown below:
(My solution) Take any $\varepsilon \in (0, 1)$. Now $f_n(x)$ denotes the integrand of the integral in question ($f_n(x)$ depends on $m$ of course, but $m$ is fixed in this problem so no problems may occur even if $m$ is not appear in the notation). First, split the integral into two pieces, $$ \int_1^e f_n(x) dx = \int_1^{e - \varepsilon} f_n(x) dx + \int_{e - \varepsilon}^e f_n(x) dx, $$ and call the first term $I_1$ and the second $I_2$. Since all of $x^m$, $e^x$ and $(\log x)^n$ are increasing functions on $[1, e]$, so is $f_n(x)$. Then, $I_1$ is evaluated as below: $$ I_1 \le \int_1^{e - \varepsilon} f_n(e - \varepsilon) dx = f_n(e - \varepsilon)(e - \varepsilon - 1). $$ Similarly, $I_2$ is evaluated like $$ I_2 \le \int_{e - \varepsilon}^e f_n(e) = f_n(e)\varepsilon. $$ Since $\log(e - \varepsilon) \in (0, 1)$, I obtained $$ \lim_{n \to \infty} f_n(e - \varepsilon)(e - \varepsilon - 1) = \lim_{n \to \infty} (e - \varepsilon)^m e^{e - \varepsilon} (\log(e - \varepsilon))^n = 0 $$ and $$ \lim_{n \to \infty} f_n(e)\varepsilon = \lim_{n \to \infty} e^m e^e (\log e)^n \varepsilon = e^{m + e}\varepsilon. $$ Therefore, I conclude that $$ 0 \le \liminf_{n \to \infty} \int_1^e f_n(x) dx \le \limsup_{n \to \infty} \int_1^e f_n(x) \le \limsup_{n \to \infty} (I_1 + I_2) \le \limsup_{n \to \infty} (f_n(e - \varepsilon)(e - \varepsilon - 1) + f_n(e)\varepsilon) = e^{m + e}\varepsilon. $$ Since $\varepsilon$ is arbitrary, both $\liminf$ and $\limsup$ of the integral as $n \to \infty$ must be $0$, which indicates $$ \lim_{n \to \infty} \int_1^e x^m e^x (\log x)^n dx = 0. $$
Is my proof valid or invalid? If there are either trivial or nontrivial mistakes, please let me know which part I should modify. Especially, my most worried thing is how to use $\liminf$ and $\limsup$.
Your proof is valid, and very well-written.
Eventually you will learn about Measure Theory and Dominated Convergence, and then this one is a one-liner!