Question about $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$

Question

$\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$

Let $x=t^6$, then it becomes $\lim\frac{t^2 - (\sqrt[6]{2})^2}{t^3 - (\sqrt[6]{2})^3}$=$\lim\frac{(t+\sqrt[6]{2})(t-\sqrt[6]{2})}{(t-\sqrt[6]{2})(t^2+t\sqrt[6]{2}+(\sqrt[6]{2})^2}$=$\lim\frac{(t+\sqrt[6]{2})}{(t^2+t\sqrt[6]{2}+(\sqrt[6]{2})^2}$

But when $x\to 2$, t can go to $\sqrt[6]{2}$ or -$\sqrt[6]{2}$, which gives two different limits, where I was wrong? Thanks!

Answer 1

You can only use $t$ going to $\sqrt[6]{2}$. This is because you have $t^3 = \sqrt{x} \gt 0$ for $x \to 2$, but $t = -\sqrt[6]{2}$ means $t^3 \lt 0$.

Answer 2

We can write your idea in the following form.

Let $\sqrt[6]{\frac{x}{2}}=y.$

Thus, $$\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}=\frac{\sqrt[3]2}{\sqrt2}\lim_{x\rightarrow2}\frac{\sqrt[3]{\frac{x}{2}}-1}{\sqrt{\frac{x}{2}}-1}=\frac{\sqrt[3]2}{\sqrt2}\lim_{y\rightarrow1}\frac{y^2-1}{y^3-1}=$$ $$=\frac{\sqrt[3]2}{\sqrt2}\lim_{y\rightarrow1}\frac{y+1}{y^2+y+1}=\frac{\sqrt[3]2}{\sqrt2}\cdot\frac{2}{3}=\frac{\sqrt[6]{32}}{3}.$$ I think, in this form just impossible to make a mistake.

Answer 3

The common method is to complete: $$\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}\cdot \frac{\sqrt[3]{x^2}+\sqrt[3]{2x}+\sqrt[3]{2^2}}{\sqrt[3]{x^2}+\sqrt[3]{2x}+\sqrt[3]{2^2}}\cdot \frac{\sqrt{x} + \sqrt{2}}{\sqrt{x} + \sqrt{2}}=\\ \lim\limits_{x \to2} \frac{\sqrt{x} + \sqrt{2}}{\sqrt[3]{x^2}+\sqrt[3]{2x}+\sqrt[3]{2^2}}=\\ \frac{2\sqrt{2}}{3\sqrt[3]{4}}=\frac13\cdot 2^{5/6}.$$

Answer 4

Let $x=y+2$ $$\frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}=\frac{\sqrt[3]{y+2} - \sqrt[3]{2}}{\sqrt{y+2} - \sqrt{2}}$$ and use the binomial theorem or Taylor series around $y=0$. Using Taylor, you would have $$\frac{\frac{y}{3\ 2^{2/3}}-\frac{y^2}{18\ 2^{2/3}}+O\left(y^3\right) } {\frac{y}{2 \sqrt{2}}-\frac{y^2}{16 \sqrt{2}}+O\left(y^3\right) }$$ Divide top and bottom by $y$ and long division to get $$\frac{2^{5/6}}{3}-\frac{y}{36 \sqrt[6]{2}}+O\left(y^2\right)$$ which shows the limit and also how it is approached.

Answer 5

Use the (not so) popular algebraic limit formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ Dividing both the numerator and denominator of the expression under limit by $x-2$ we can see that the desired limit equals $$\left. \frac{1}{3}\cdot 2^{-2/3}\middle/\frac{1}{2}\cdot 2^{-1/2}\right.=\frac {2^{5/6}}{3}$$