<s>[INST] Find the limit $\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x}$ I am trying to evaluate the following limit
$$\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x}$$ 
If we use the binomial expansion of the numerator term, the answer is $\frac{1}{4}$. The same answer is obtained if we apply L'Hospital's rule.
Is it possible to solve this using something even more basic ? i.e. without recourse to binomial expansion or L'Hospital's rule or definition of derivative ? The reason I ask this is because this problem (from a textbook) comes up before the introduction of L'Hospital's rule.

user37238 wanted to know how the binomial expansion can be used.Basically, we have
$(1+x)^n= 1 + nx + \frac{n (n-1)}{2} + \frac{n (n-1)(n-2)}{2!} + \ldots$
Similarly, 
$(1-x)^n= 1 - nx + \frac{n (n-1)}{2} - \frac{n (n-1)(n-2)}{2!} + \ldots$
Using these expansions with $n=\frac{1}{8}$ in the given problem gives the answer. 
 [/INST] For example, use
\begin{align}
&\;\;\;\frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8}}{x}=\\&=\frac{1}{x}\cdot\frac{(1+x)^\frac{1}{4}-(1-x)^\frac{1}{4}}{(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}}=\\
&=\frac{1}{x}\cdot\frac{(1+x)^\frac{1}{2}-(1-x)^\frac{1}{2}}{\left[(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}\right]\left[(1+x)^\frac{1}{4}+(1-x)^\frac{1}{4}\right]}=\\
&=\frac{1}{x}\cdot\frac{(1+x)-(1-x)}{\left[(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}\right]\left[(1+x)^\frac{1}{4}+(1-x)^\frac{1}{4}\right]\left[(1+x)^\frac{1}{2}+(1-x)^\frac{1}{2}\right]}=\\
&=\frac{2}{\left[(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}\right]\left[(1+x)^\frac{1}{4}+(1-x)^\frac{1}{4}\right]\left[(1+x)^\frac{1}{2}+(1-x)^\frac{1}{2}\right]}.
\end{align}
Now to compute the limit, it suffices to set $x=0$ in the last expression.
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