Find $\lim\limits_{x\to 0} \frac{\sqrt{\ 1+x} - \sqrt{\ 1-x}}{\sqrt[3]{\ 1+x} - \sqrt[3]{\ 1-x}}$ This problem 
$$\lim_{x\to 0} \frac{\sqrt{\ 1+x} - \sqrt{\ 1-x}}{\sqrt[3]{\ 1+x} - \sqrt[3]{\ 1-x}}$$
is from Silverman's "Modern Calculus and Analytical Geometry" Section 22, #16d. I've been struggling on it for a while and can't figure out what to do besides trying to multiply by the conjugate and/or substitution but it doesn't work out. What do you all think? Keep in mind you can't use L'Hopital's rule, only elementary math.

Added for your curiosity.
For the limit itself, you aleary received good answers. The problem can also be addressed using Taylor series or the generalized binomial theorem
$$(1+x)^a=1+a x+\frac{1}{2} a(a-1)  x^2+\frac{1}{6} a(a-1) (a-2)  x^3+O\left(x^4\right)$$ making
$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+O\left(x^4\right)$$
$$\sqrt{1-x}=1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}+O\left(x^4\right)$$
$$\sqrt[3]{1+x}=1+\frac{x}{3}-\frac{x^2}{9}+\frac{5 x^3}{81}+O\left(x^4\right)$$
$$\sqrt[3]{1-x}=1-\frac{x}{3}-\frac{x^2}{9}-\frac{5 x^3}{81}+O\left(x^4\right)$$
$$ \frac{\sqrt{\ 1+x} - \sqrt{\ 1-x}}{\sqrt[3]{\ 1+x} - \sqrt[3]{\ 1-x}}=\frac{ x+\frac{1}{8}x^3+O\left(x^4\right)} {\frac{2 }{3}x+\frac{10 }{81}x^3+O\left(x^4\right) }=\frac{3}{2}-\frac{13 x^2}{144}+O\left(x^3\right)$$ which shows the limit and how it is approached.
But what is interesting is that this is a very good approximation even if $x$ is not small. For example, using $x=\frac 12$, the exact expression is $\approx 1.47469$ while the approximation leads to $\frac{851}{576}\approx 1.47743$.
Edit
Making the problem more general and using the same steps
$$\frac{(x+1)^p-(1-x)^p}{(x+1)^q-(1-x)^q}=\frac{p}{q}+\frac{p  (p-q) (p+q-3)}{6 q}x^2+O\left(x^4\right)$$
