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. 

$$\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x} =\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-1+1-(1-x)^\frac{1}{8} }{x}=\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-1 }{x}+\lim_{x \rightarrow 0} \frac{(1-x)^\frac{1}{8}-1 }{-x}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4} $$
We applied the known limit for $a\in R$:
$$\lim_{x \rightarrow 0} \frac{(1+x)^a-1 }{x}=a $$
