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

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 $t^3 \lt 0$.
