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. 
 A: 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.
A: Written in the form $2\cdot \frac{(1+x)^\frac18-(1-x)^\frac18}{2x},$ the expression is twice the difference quotient for the derivative at $t=1$ of $f(t)=t^{1/8}.$ It's actually a "symmetric" difference quotient, which for a differentiable function will approach the usual derivative. So the limit is $2\cdot f'(1)=\frac14.$
Note that then this problem could be done just from the definition of the derivative, provided one is OK with using the symmetric difference quotient, and so does not involve L'Hospital or binomial expansion.
A: $$\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x} =\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-1+1-(1-x)^\frac{1}{8} }{x}=\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-1 }{x}+\lim_{x \rightarrow 0} \frac{(1-x)^\frac{1}{8}-1 }{-x}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4} $$
We applied the known limit for $a\in R$:
$$\lim_{x \rightarrow 0} \frac{(1+x)^a-1 }{x}=a $$
A: Hint
When $x$ is small compared to $1$, $(1+x)^n \simeq 1+nx$. This is a property you must remember.
I am sure that you can take from here.
