Elementary way to evaluate $\lim_{x\to0}\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x}$ 
Evaluate:
  $$
\lim_{x\to0}\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x},\ a>0,\ n\in\Bbb N
$$

I've given it several tries but couldn't find an elementary method to find the limit. Two other ways that worked are L'Hospital's rule and generalized binomial expansion.

First method:
$$
\lim_{x\to0} f(x) = \lim_{x\to0}\frac{g(x)}{h(x)} = \lim_{x\to0}\frac{g'(x)}{h'(x)} \\ = \lim_{x\to0}\left({1\over n}\left(a+x\right)^{{1\over n} -1} + {1\over n}(a-x)^{{1\over n} - 1}\right) = {2\over n}{a^{n-1\over n}} = \frac{2\sqrt[n]{a}}{na}
$$

Second method:
$$
(a+x)^{1\over n} = \sqrt[n]{a}\left(1 + {x\over a}\right)^{1\over n} =\\
\sqrt[n]{a}\left(1 + {1\over n}{x\over a} + \frac{\left({1\over n}\right)\left({1\over n} - 1\right)}{2!}\left({x\over a}\right)^2 + \cdots \right)
$$
Also:
$$
(a-x)^{1\over n} = \sqrt[n]{a}\left(1 - {x\over a}\right)^{1\over n} = \\
\sqrt[n]{a}\left(1 - {1\over n}{x\over a} + \frac{\left({1\over n}\right)\left({1\over n} - 1\right)}{2!}\left({x\over a}\right)^2 + \cdots \right)
$$
Combining those ones may obtain:
$$
\lim_{x\to0}f(x) = \lim_{x\to0}\frac{\sqrt[n]{a}\left({2x\over na} + O(x^2)\right)}{x} = \frac{2\sqrt[n]{a}}{na}
$$

The problem is I'm not supposed to use derivatives for solving that limit. Also, generalized binomial expansion is somewhat too complicated as well. 
Are there any elementary methods to evaluate the limit from the problem section?
I've also been trying to cast the expression to the form:
$$
\lim_{x\to a}\frac{x^n - a^n}{x - a} = na^{n-1}
$$
but failed.

Just write


*

*$\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x} = 2 \frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{(a+x) - (a-x)}$
Now, use $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$
So, you get
\begin{eqnarray*} \frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x} 
& = & 2 \frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{(a+x) - (a-x)}\\
& = & \frac{2}{\sum_{k=0}^{n-1}\sqrt[n]{(a+x)^{n-1-k}(a-x)^k}}\\
& \stackrel{x \to 0}{\longrightarrow} & \frac{2}{\sum_{k=0}^{n-1}\sqrt[n]{a^{n-1-k}a^k}} \\
& = & \frac{2}{n\sqrt[n]{a^{n-1}}} \\
& = & \frac{2\sqrt[n]{a}}{na} \\
\end{eqnarray*}
