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?


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 $$\displaystyle \lim_{t\rightarrow 2}\frac{t^2+2t+4}{t+2} =\frac{12}{4} = 3$$


Another approach is to use the binomial expansion


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

$$\begin{align} \sqrt{x}-8&=8\left(\left(1+\frac{x-64}{64}\right)^{1/2}-1\right)\\\\ &=8\sum_{k=1}^{\infty}\binom{1/2}{k}\left(\frac{x-64}{64}\right)^k\\\\ &=\frac{8}{2}\frac{x-64}{64}\left(1+O(x-64)\right)\tag 1 \end{align}$$


$$\begin{align} \sqrt[3]x-4&=4\left(\left(1+\frac{x-64}{64}\right)^{1/3}-1\right)\\\\ &=4\sum_{k=1}^{\infty}\binom{1/3}{k}\left(\frac{x-64}{64}\right)^k\\\\ &=\frac{4}{3}\frac{x-64}{64}\left(1+O(x-64)\right)\tag 2 \end{align}$$

Taking the ratio of $(1)$ and $(2)$ and passing to the limit gives the result $3$ as expected.


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 $$\lim_{x \to 64} \frac{\sqrt{x}-8}{\sqrt[3]{x}-4} = \lim_{x \to 64} \frac{\sqrt[3]{x^2} + 4 \sqrt[3]{x} + 16}{\sqrt x + 8} = 3.$$



Let $x^{1/6}=t$. Then, $\sqrt x=t^3,\sqrt[3]{x}=t^2$.

So, $$\frac{\sqrt{x}-8}{\sqrt[3]{x}-4}=\frac{t^3-8}{t^2-4}=\frac{\color{red}{(t-2)}(t^2+2t+4)}{\color{red}{(t-2)}(t+2)}$$

  • $\begingroup$ Thanks! I can see it now. <3 $\endgroup$ – Mark Sep 14 '15 at 15:42
  • $\begingroup$ @MarkHC: You are welcome. (By the way, the "6" comes from $lcm(2,3)=6$.) $\endgroup$ – mathlove Sep 14 '15 at 15:48

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