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 $a=(x+1)^{1/6}$ and $b=(x-1)^{1/6}$. Then the expression becomes $\lim_{x→0}\frac{a^3-b^3}{a^2-b^2}$. This does divide through. It is equivalent to $\lim_{x→0}\frac{a^2+ab+b^2}{a+b}=3/2$.
Let $$a=(x+1)^{\frac{1}{6}}$$ $$b=(1-x)^{\frac{1}{6}}$$ Our equation becomes, $$\lim_{x\to 0} \frac{a^3-b^3}{a^2-b^2}$$ $$\lim_{x\to 0} \frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}$$
$$\lim_{x\to 0} \frac{a^2+ab+b^2}{a+b}$$$$\lim_{x\to 0} \frac{(x+1)^{\frac{1}{3}}+(1-x^2)^{\frac{1}{6}}+(1-x)^{\frac{1}{3}}}{(x+1)^{\frac{1}{6}}+(1-x)^{\frac{1}{6}}}$$$$\frac{1+1+1}{1+1}$$$$\frac{3}{2}$$
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 $$\frac{(x+1)^p-(1-x)^p}{(x+1)^q-(1-x)^q}=\frac{p}{q}+\frac{p (p-q) (p+q-3)}{6 q}x^2+O\left(x^4\right)$$
\begin{align} & \lim_{x\to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt[3]{1+x} - \sqrt[3]{1-x}} = \lim_{u,v\to1} \frac{u^3-v^3}{u^2-v^2} \\[10pt] = {} & \lim_{u,v\to1} \frac{(u-v)(u^2+vu+v^2)}{(u-v)(u+v)} = \lim_{u,v\to1} \frac{u^2+uv+v^2}{u+v} = \frac 3 2. \end{align}
Here of course $u = \sqrt{\sqrt[3]{1+x}} = \sqrt[6]{1+x}$ and $v=\sqrt{\sqrt[3]{1-x}}=\sqrt[6]{1-x}.$ An obvious qualm about this is that $\lim\limits_{u,v\to1}$ ought to mean a limit as the point $(u,v)$ approaches $(1,1)$ in the plane, whereas the two radicals do not approach $1$ independently of each other, but rather the pair moves along a particular curve. However, since we ultimately see that it does not matter along which curve $(u,v)$ approaches $(1,1),$ it follows that that does not alter the bottom line.
$$\lim_{x\to 0}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}} = \lim_{x\to 0}\frac{\sqrt[3]{(1+x)^2}+\sqrt[3]{1-x^2} +\sqrt{(1-x)^2}}{\sqrt{1+x}+\sqrt{1-x}} = \frac{3}{2}.$$
