<s>[INST] 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.
 [/INST] 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
$$\frac{(x+1)^p-(1-x)^p}{(x+1)^q-(1-x)^q}=\frac{p}{q}+\frac{p  (p-q) (p+q-3)}{6 q}x^2+O\left(x^4\right)$$
</s>