Evaluate a limit involving a definite integral 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!
 A: Hint: Expand the integrand by using the geometric series:
$$\frac{x^n}{5(1+4x/5)}.$$
It is uniformly convergent on $|x|\leq 5/4$ and integrate the resulting polynomial.
A: HINT:
If $\displaystyle I_m=\int_0^1\dfrac{x^m}{4x+5}dx,$
$\displaystyle4I_{n+1}=\int_0^1\dfrac{4x^{n+1}}{4x+5}dx=\int_0^1\dfrac{x^n(4x+5)-5x^n}{4x+5}dx=\int_0^1x^n\ dx-5I_n$
Now $\displaystyle4\cdot\dfrac{I_{n+1}}{n+1}+\dfrac{5n}{n+1}\cdot\dfrac{I_n}n=\dfrac{\int_0^1x^n\ dx}{n+1}$
Finally,  $$\lim_{n\to\infty}\dfrac{I_{n+1}}{n+1}=\lim_{n\to\infty}\dfrac{I_n}n$$ 
A: Expand the power series:

$$I_n=\int_0^1{1\over 5}\sum_{i=0}^\infty (-1)^i\left({4\over 5}\right)^ix^{n+i}$$

Since the function converges uniformly on $[0,1]$ and absolutely, we can pull the integral inside and evaluate and we get
$$I_n = {1\over 5}\sum_{i=0}^\infty(-1)^i\left({4\over 5}\right)^i{1\over n+i+1}$$
Multiply by $n$ and we get
$$nI_n = {1\over 5}\sum_{i=0}^n(-1)^{i}\left({4\over 5}\right)^{i}{n\over n+i+1}={1\over 5}\sum_{i=0}^\infty (-1)^{i}\left({4\over 5}\right)^{i}\left(1-{i+1\over n+i+1}\right)$$
We can evaluate the first sum easily enough
$${1\over 5}\sum_{i=0}^\infty (-1)^{i}\left({4\over 5}\right)^i={1\over 5+4}={1\over 9}$$
Now for the second term we can use absolute convergence to take the limit inside the sum and we get:
$$\lim_{n\to\infty} \sum_{i=0}^\infty(-1)^i\left({4\over 5}\right)^i{i+1\over n+i+1}=0=\sum_{i=0}^\infty(-1)^i\left({4\over 5}\right)^i(i+1)\cdot\lim_{n\to\infty}{1\over n+i+1}=0$$
so the end result is just ${1\over 9}$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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$\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}
A: It is enough to apply integration by parts:
$$ n I_n = \int_{0}^{1}(n x^{n-1})\frac{x\,dx}{4x+5} = \color{red}{\frac{1}{9}}-\int_{0}^{1}\frac{5x^n}{(4x+5)^2}\,dx$$
and notice that:
$$ \left|\int_{0}^{1}\frac{5x^n}{(4x+5)^2}\,dx\right|\leq \frac{1}{5}\int_{0}^{1}x^n\,dx = \frac{1}{5n+5}.$$
