Elementary way to evaluate $\lim_{x\to0}\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x}$

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*}

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.

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 $$\frac{1}{n}a^{(1/n)-1}+\frac{1}{n}a^{(1/n)-1}=\frac{2a^{1/n}}{na}$$