Evaluate a limit involving a definite integral

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: $$ \left|\int_{0}^{1}\frac{5x^n}{(4x+5)^2}\,dx\right|\leq \frac{1}{5}\int_{0}^{1}x^n\,dx = \frac{1}{5n+5}.$$

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.

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, $$\lim_{n\to\infty}\dfrac{I_{n+1}}{n+1}=\lim_{n\to\infty}\dfrac{I_n}n$$