Alternative solutions to $\lim_{n\to\infty} \frac{1}{\sqrt{n}}\int_{ 1/{\sqrt{n}}}^{1}\frac{\ln(1+x)}{x^3}\mathrm{d}x$ Here is a limit that can be computed directly by performing the integration and then taking
the limit, but the way is rather ugly. What else can we do? Might we avoid the integration? 
$$\lim_{n\to\infty} \frac{1}{\sqrt{n}}\int_{ 1/{\sqrt{n}}}^{1}\frac{\ln(1+x)}{x^3}\mathrm{d}x$$
 A: By L`Hopital's rule:
$$ \lim_{n\rightarrow \infty} -\frac{\ln(1 + \sqrt{n}) n^{1.5} \frac{-1}{2 n^{1.5}}}{0.5 n^{-0.5}}$$
$$ = \lim_{n\rightarrow \infty} \ln(1 + \frac{1}{\sqrt{n}}) \sqrt{n}$$
Using the fact that
$$\ln(1 + x) = x + O(x^2)$$
We see that the limit is $1$.
A: $$\lim_{n\to \infty}\dfrac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1\dfrac{\ln(1+x)}{x^3}\mathrm dx=\lim _{t\to \infty}\frac{1}{t}\int_{\frac{1}{t}}^1\dfrac{\ln(1+x)}{x^3}\mathrm dx= \lim _{t\to \infty}\ln(1+t^{-1})t^3/t^2=1$$ By l'Hospital's rule and $\log(1+x)/x\to1$
A: Not strictly avoiding the integration, but expanding the integrand makes it pretty straightforward:
$$\frac{\ln(1+x)}{x^3}=\frac{1}{x^2}-\frac{1}{2x}+\frac{1}{3}-\frac{x}{4}+\cdots$$
$$\int_{\frac{1}{\sqrt{n}}}^1 \frac{\ln (1+x)}{x^3}dx=\left[-\frac{1}{x}-\frac{\ln x}{2}+\frac{x}{3}-\cdots\right]_{1/\sqrt{n}}^1=-\frac{3}{4}+\sqrt{n}+\frac{\ln \sqrt{n}}{2}-\frac{1}{3\sqrt{n}}+\cdots$$
$$\frac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1 \frac{\ln (1+x)}{x^3}dx=1-\frac{3}{4\sqrt{n}}+\frac{\ln \sqrt{n}}{2\sqrt{n}}-\frac{1}{3n}+\cdots\to 1$$
A: Close to $x = 0$, we have
$$\frac{\log(1+x)}{x^3} \sim \frac{1}{x^2} - \frac{1}{2x} + \cdots$$
Pick a $\delta > 0$ such that the error term in R.H.S is $O(1)$ on $(0,\delta)$, we have:
$$\begin{align} \int_{\frac{1}{\sqrt{n}}}^1 \frac{\log(1+x)}{x^3} dx 
&= \left(\int_{\frac{1}{\sqrt{n}}}^\delta + \int_{\delta}^1\right) \frac{\log(1+x)}{x^3} dx\\
&= \left[ -\frac{1}{x} -\frac12 \log x \right]_{\frac{1}{\sqrt{n}}}^\delta + O(\delta) + \int_{\delta}^1 \frac{\log(1+x)}{x^3} dx\\
&= \sqrt{n} +\frac12 \log ( \sqrt{n} ) + O(\frac{1}{\delta})
\end{align}$$
Notice $\delta$ has be chosen independent of $n$, we have:
$$\lim_{n\to\infty} \frac{1}{\sqrt{n}} \int_{\frac{1}{\sqrt{n}}}^1 \frac{\log(1+x)}{x^3} dx
= \lim_{n\to\infty} \frac{1}{\sqrt{n}} \left(\sqrt{n} +\frac12 \log ( \sqrt{n} ) + O(\frac{1}{\delta})\right) = 1$$
A: We have with the fact $x-\frac{x^2}{2}\leq\log(1+x)\leq x$,
$$\lim_{n\to \infty}\frac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1\frac{\log(1+x)}{x^3}\leq\lim_{n\to \infty}\frac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1\frac{1}{x^2}=\lim_{n\to \infty}\frac{1}{\sqrt{n}}({\sqrt{n}}-1)=1.$$
and
$$1=\lim_{n\to \infty}\frac{1}{\sqrt{n}}(({\sqrt{n}}-1)-\frac{1}{2}\log\sqrt{n})=\lim_{n\to \infty}\frac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1\frac{1}{x^2}-\frac{1}{2x}\leq\lim_{n\to \infty}\frac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1\frac{\log(1+x)}{x^3}$$
We can conclude.
A: An alternate way is to notice that expanding the integrand also gives:
$$\frac{1}{x^2}-\frac{1}{x}<\frac{\ln(1+x)}{x^3}< \frac{1}{x^2}\;\,\text{for}\;\,0<x<1$$ 
Hence:
$$\frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{1}{x^2}-\frac{1}{x}\,dx< \frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{\ln(1+x)}{x^3}dx< \frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{1}{x^2}dx$$ 
$$1-\frac{1}{\sqrt{n}}-\frac{\ln\sqrt{n}}{\sqrt{n}}<\frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{\ln(1+x)}{x^3}dx< 1-\frac{1}{\sqrt{n}}$$ 
$$\frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{\ln(1+x)}{x^3}\to 1$$
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\lim_{n \to \infty}{1 \over \root{n}}
     \int_{ 1/\root{n}}^{1}{\ln\pars{1 + x} \over x^{3}}\,\dd x}$

\begin{align}
&\color{#00f}{\large\lim_{n \to \infty}\bracks{{1 \over \root{n}}%
\int_{ 1/\root{n}}^{1}{\ln\pars{1 + x} \over x^{3}}\,\dd x}}
\\[3mm]&=\lim_{n \to \infty}\bracks{{1 \over \root{n}}%
\int_{ 1/\root{n}}^{1}{\ln\pars{1 + x} - x + x^{2}/2 \over x^{3}}\,\dd x
+{1 \over \root{n}}%
\int_{ 1/\root{n}}^{1}{x - x^{2}/2 \over x^{3}}\,\dd x}
\\[3mm]&=\lim_{n \to \infty}\bracks{{1 \over \root{n}}%
\int_{ 1/\root{n}}^{1}{\ln\pars{1 + x} - x + x^{2}/2 \over x^{3}}\,\dd x
+ 1 - {1 \over \root{n}} - {\ln\pars{n} \over 4\root{n}}}
\\[3mm]&=1 + \lim_{n \to \infty}\bracks{{1 \over \root{n}}%
\int_{ 1/\root{n}}^{1}{\ln\pars{1 + x} - x + x^{2}/2 \over x^{3}}\,\dd x - {1 \over \root{n}} - {\ln\pars{n} \over 4\root{n}}}
=\color{#00f}{\Large 1}
\end{align}

