<s>[INST] 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!
 [/INST] 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.
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