TITLE: Evaluate a limit involving a definite integral
QUESTION [22 upvotes]: Let $(I_n)_{n \geq 1}$ be a sequence such that: 
$$I_n = \int_0^1 \frac{x^n}{4x + 5} dx$$ 
Evaluate the following limit:
$$\lim_{n \to \infty} nI_n$$
All I've been able to find is that $(I_n)$ is decreasing and converges to $0$.
Thank you!

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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$\ds{\mrm{g}\pars{n,x} \equiv {\pars{1 - x}^{n} \over 9 - 4x}}$


This is an application of Laplace Method:
\begin{align}
\lim_{n \to \infty}\pars{n\int_{0}^{1}{x^{n} \over 4x + 5}\,\dd x} & =
{1 \over 9}\lim_{n \to \infty}\bracks{n
\int_{0}^{1}{\pars{1 - x}^{n} \over 1 - 4x/9}\,\dd x} =
{1 \over 9}\lim_{n \to \infty}\pars{n\int_{0}^{\infty}\expo{-nx}\,\dd x}
\\[5mm] & =
\bbx{\ds{1 \over 9}}
\end{align}