Proving the limit

$$\dfrac{n}{\sum_i \dfrac{1}{x+a_i}} - x = \dfrac{n-\sum_j\dfrac{x}{x+a_j}}{\sum_i \dfrac{1}{x+a_i}} = \dfrac{n-\sum_j\dfrac{x+a_j}{x+a_j}+\sum_j\dfrac{a_j}{x+a_j}}{\sum_i \dfrac{1}{x+a_i}} = \dfrac{\sum_j\dfrac{a_j}{x+a_j}}{\sum_i \dfrac{1}{x+a_i}} =$$ $$= \sum_j \dfrac{a_j}{\sum_i\dfrac{x+a_j}{x+a_i}}$$

Now $$\displaystyle\lim_{x\to\infty} \sum_j \dfrac{a_j}{\sum_i\dfrac{x+a_j}{x+a_i}} = \sum_j \dfrac{a_j}{\sum_i\displaystyle\lim_{x\to\infty}\dfrac{x+a_j}{x+a_i}} = \sum_j \dfrac{a_j}{n} = \dfrac{a_i + \ldots + a_n}{n}$$

You may write $1/(x+a_1) = 1/x \cdot 1/(1+a_1/x) = 1/x \left(1-a_1/x+O\left(1/x^2\right)\right)$.

Here $g(x)=O(f(x))$ means $g(x)/f(x)$ is bounded as $x\rightarrow \infty$.

So

$$\begin{align}\lim_{x\to \infty}\left(\frac{n}{\sum_{i=1}^n{\frac{1}{x+a_i}}}-x\right) &=\lim_{x\to \infty}\left(\frac{n}{\frac{1}{x}\sum_{i=1}^n\left({1-\frac{a_i}{x}+O\left(1/x^2\right)}\right)}-x\right)\\ &=\lim_{x\to \infty}\left(\frac{n}{\frac{1}{x}\left(n-\frac{\sum a_i}{x}+O(1/x^2)\right)}-x\right)\\ &=\lim_{x\to \infty}\left(x\frac{1}{\left(1-\frac{\sum a_i}{nx}+O(1/x^2)\right)}-x\right)\\ &=\lim_{x\to \infty}\left(x\left(1+\frac{\sum a_i}{nx}+O(1/x^2)\right)-x\right)\\ &=\lim_{x\to \infty}\left(\frac{\sum a_i}{n}+O(1/x)\right)\\ &=\frac{a_1+\cdots+a_n}{n} \end{align}$$

Hints: Factor out $1/x$ downstairs and use

$$\frac{1}{1+u} = 1-u +\frac{u^2}{1+u}.$$