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.
 A: 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*}
A: First it should be easy enough to see that your limit is
$$ 2 \lim_{x\to 0}\frac{\sqrt[n]{a+x}-\sqrt[n]{a}}{x} $$
Now switch variables to $y=\sqrt[n]{a+x}$, giving $x=y^n-a$:
$$ \cdots = 2 \lim_{y\to b}\frac{y-b}{y^n-b^n} \qquad\text{where }b=\sqrt[n]{a}$$
Take the limit of the reciprocal instead:
$$ \cdots = \frac{2}{\lim\limits_{y\to b}\frac{y^n-b^n}{y-b}} $$
and you can now use the rule that you write you already have available.
A: The last formula in your question is the key here. Let $u=a+x, v=a-x$ so that both $u, v$ tend to $a$ as $x\to 0$. Also $(u-a) /x\to 1$ and $(v-a) /x\to - 1$. The given expression can be written as $$\frac{u^{1/n}-a^{1/n}}{u-a}\cdot\frac{u-a}{x}-\frac{v^{1/n}-a^{1/n}}{v-a}\cdot\frac{v-a}{x}$$ which tends to $$\frac{1}{n}a^{(1/n)-1}+\frac{1}{n}a^{(1/n)-1}=\frac{2a^{1/n}}{na}$$
A: With $$\sqrt[n]{a+x}=u,\\\sqrt[n]{a-x}=v,$$
$$\frac{u-v}x=\frac{u^n-v^n}{x(u^{n-1}+u^{n-2}v+\cdots v^{n-1})}=\frac2{u^{n-1}+u^{n-2}v+\cdots v^{n-1}}.$$
Every term in the denominator tends to $a^{(n-1)/n}$ and there are $n$ of them.

Alternatively, let $x=at$ and pull $a$ out of the expression,
$$\frac{\sqrt[n]{a+at} - \sqrt[n]{a-at}}{at}=a^{1/n-1}\frac{\sqrt[n]{1+t} - \sqrt[n]{1-t}}{t},$$
where the fraction tends to a constant. This brings you halfway of the work.
Now,
$$\lim_{t\to0}\frac{\sqrt[n]{1+t} - 1+1-\sqrt[n]{1-t}}{t}=\left.\left(\left(\sqrt[n]{1+t}\right)'-\left(\sqrt[n]{1-t}\right)'\right)\right|_{t=0}
=2\left.\left(\sqrt[n]{1+t}\right)'\right|_{t=0}.$$
You can compute the derivative or use the Taylor development of$\sqrt[n]{1+t}$, and establish that the constant is $\dfrac2n$.
A: Alternatively:
$$\lim_{x\to0}\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x}=
\lim_{x\to0}\frac{\sqrt[n]{a-x}\cdot \left(\sqrt[n]{\frac{a+x}{a-x}}-1\right)}{x}\stackrel{\frac{a+x}{a-x}=t^n}{=}\\
\sqrt[n]{a}\cdot \lim_{t\to1}\frac{t-1}{\frac{a(t^n-1)}{t+1}}=
\frac{2\sqrt[n]{a}}{a}\cdot\lim_{t\to1}\frac{ t-1}{t^n-1}=\frac{2\sqrt[n]{a}}{an}.$$
A: Rewrite the expression as
$$\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x}=\frac{\sqrt[n]{a+x} - \sqrt[n]{a}}{x}+\frac{\sqrt[n]{a-x} - \sqrt[n]{a}}{-x}$$
and observe each fraction is a rate of variation from $a$. hence the limit  of each is the derivative at $a$.
