Limits 2

The first equality is fine.

However, you cannot randomly interchange the limits with the integration sign. In particular, you need to justify the claim that:

$\lim_{ n\rightarrow \infty} \int_0^1 \frac{ x\, dx} { 1 + \frac{1}{n} + \frac{x}{n} } = \int_0^1 \lim_{n \rightarrow \infty} \frac{x\, dx } { 1 + \frac{1}{n} + \frac{x}{n} }.$

In order to justify the interchange of limits (because integration is essentially a limit calculation), you need to apply a theorem like dominated convergence theorem or monotone convergence theorem .

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We can approach this step by taking the integral and applying the limits:

$\lim_{ n\rightarrow \infty} \int_0^1 \frac{ x\, dx} { 1 + \frac{1}{n} + \frac{x}{n} } = \lim_{ n \rightarrow \infty} n [ (n+1) \log (n+1) - (n+1) \log(n+2) + 1 ]$

Another approach:
$\displaystyle\lim_{n\to\infty}\sum_{r=1}^n \frac{r}{n^2+n+r}=\lim_{n\to\infty}\frac 1n\sum_{r=1}^n \frac{\frac rn}{1+\frac 1n+\frac r{n^2}}=\int_0^1 \frac x1 \mathrm{d}x=\frac 12$