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?
 A: Let $x=t^6\;,$ Then when $x\rightarrow 64\;,$ Then $t\rightarrow 2$
So limit Convert into $$\displaystyle \lim_{t\rightarrow 2}\frac{t^3-8}{t^2-4} = \lim_{t\rightarrow 2}\frac{(t-2)\cdot (t^2+2t+4)}{(t-2)(t+2)}$$
So we get $$\displaystyle \lim_{t\rightarrow 2}\frac{t^2+2t+4}{t+2} =\frac{12}{4} = 3$$
A: Another approach is to use the binomial expansion 
$$(1+x)^r=\sum_{k=0}^{\infty}\binom{r}{k}x^k$$
where the binomial coefficient for non-integer $r$ is defined by 
$$\binom{r}{k}=\frac{r(r-1)\cdots (r-k+1)}{k!}$$
Then, we have
$$\begin{align}
\sqrt{x}-8&=8\left(\left(1+\frac{x-64}{64}\right)^{1/2}-1\right)\\\\
&=8\sum_{k=1}^{\infty}\binom{1/2}{k}\left(\frac{x-64}{64}\right)^k\\\\
&=\frac{8}{2}\frac{x-64}{64}\left(1+O(x-64)\right)\tag 1
\end{align}$$
and 
$$\begin{align}
\sqrt[3]x-4&=4\left(\left(1+\frac{x-64}{64}\right)^{1/3}-1\right)\\\\
&=4\sum_{k=1}^{\infty}\binom{1/3}{k}\left(\frac{x-64}{64}\right)^k\\\\
&=\frac{4}{3}\frac{x-64}{64}\left(1+O(x-64)\right)\tag 2
\end{align}$$
Taking the ratio of $(1)$ and $(2)$ and passing to the limit gives the result $3$ as expected.
A: 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 $$\lim_{x \to 64} \frac{\sqrt{x}-8}{\sqrt[3]{x}-4} = \lim_{x \to 64} \frac{\sqrt[3]{x^2} + 4 \sqrt[3]{x} + 16}{\sqrt x + 8}  = 3.$$
A: HINT : 
Let $x^{1/6}=t$. Then, $\sqrt x=t^3,\sqrt[3]{x}=t^2$.
So,
$$\frac{\sqrt{x}-8}{\sqrt[3]{x}-4}=\frac{t^3-8}{t^2-4}=\frac{\color{red}{(t-2)}(t^2+2t+4)}{\color{red}{(t-2)}(t+2)}$$
