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.

Let
$$a=(x+1)^{\frac{1}{6}}$$
$$b=(1-x)^{\frac{1}{6}}$$
Our equation becomes,
$$\lim_{x\to 0} \frac{a^3-b^3}{a^2-b^2}$$
$$\lim_{x\to 0} \frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}$$

$$\lim_{x\to 0} \frac{a^2+ab+b^2}{a+b}$$$$\lim_{x\to 0} \frac{(x+1)^{\frac{1}{3}}+(1-x^2)^{\frac{1}{6}}+(1-x)^{\frac{1}{3}}}{(x+1)^{\frac{1}{6}}+(1-x)^{\frac{1}{6}}}$$$$\frac{1+1+1}{1+1}$$$$\frac{3}{2}$$

