Solving limit of radicals without L'Hopital $\lim_{x\to 64} \frac{\sqrt x - 8}{\sqrt[3] x - 4} $ I've been trying to solve this for some time now, but I cannot get any closer to the solution. 
I need to solve this limit without using L'Hopital's rule.
$$\lim_{x\to 64} \dfrac{\sqrt x - 8}{\sqrt[3] x - 4}  $$
By using L'Hopital I know the result should be 3, but I cannot get to it any other way...
How would I simplify this equation?

It isn't mentioned much in calculus courses, but you could rationalize the cube root in the denominator using $(A-B)(A^2 + AB + B^2) = A^3 - B^3$, to wit:
$$\frac{\sqrt{x}-8}{\sqrt[3]{x}-4} = \frac{(\sqrt{x}-8)(\sqrt x + 8)(\sqrt[3]{x^2} + 4 \sqrt[3]{x} + 16)}{(\sqrt[3]{x}-4)(\sqrt[3]{x^2} + 4 \sqrt[3]{x} + 16)(\sqrt x + 8)} = \frac{(x-64)(\sqrt[3]{x^2} + 4 \sqrt[3]{x} + 16)}{(x-64)(\sqrt x + 8)}$$ to get $$\lim_{x \to 64} \frac{\sqrt{x}-8}{\sqrt[3]{x}-4} = \lim_{x \to 64} \frac{\sqrt[3]{x^2} + 4 \sqrt[3]{x} + 16}{\sqrt x + 8}  = 3.$$
