Proving the limit I have to prove that:$$\lim_\limits{x\to\infty}\bigg(\frac{n}{\frac{1}{x+a_1}+\frac{1}{x+a_2}+\cdots+\frac{1}{x+a_n}}-x\bigg)=\frac{a_1+a_2+\cdots+a_n}{n}$$
The way I started doing this is:
$$=\lim_\limits{x\to\infty}\left(\frac{n}{\frac{(x+a_1)(x+a_2)\cdots(x+a_n)\sum_{i=1}^{n}\big(\frac{1}{x+a_i}\big)}{(x+a_1)(x+a_2)\cdots(x+a_n)}}-x\right)$$
Then I combine $x$ with the rest, but that leads me nowhere. Any tips on how to do this? Taylor expansion cannot be used.

$$\dfrac{n}{\sum_i \dfrac{1}{x+a_i}} - x =  \dfrac{n-\sum_j\dfrac{x}{x+a_j}}{\sum_i \dfrac{1}{x+a_i}} =  \dfrac{n-\sum_j\dfrac{x+a_j}{x+a_j}+\sum_j\dfrac{a_j}{x+a_j}}{\sum_i \dfrac{1}{x+a_i}} = \dfrac{\sum_j\dfrac{a_j}{x+a_j}}{\sum_i \dfrac{1}{x+a_i}} =$$ $$= \sum_j \dfrac{a_j}{\sum_i\dfrac{x+a_j}{x+a_i}}$$
Now 
$$\displaystyle\lim_{x\to\infty} \sum_j \dfrac{a_j}{\sum_i\dfrac{x+a_j}{x+a_i}} = \sum_j \dfrac{a_j}{\sum_i\displaystyle\lim_{x\to\infty}\dfrac{x+a_j}{x+a_i}} = \sum_j \dfrac{a_j}{n} = \dfrac{a_i + \ldots + a_n}{n}$$
