<s>[INST] Simple limit problem without L'Hospital's rule $$\lim_{x \to 1}\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1}$$
We are not supposed to use any derivatives yet, but I can't find any formula that helps here. It's a $\frac{0}{0}$ indeterminate form, and all I think of doing is
$$\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1} = \frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1} \cdot \frac{\sqrt{x^4 + 1} + \sqrt{2}}{\sqrt{x^4 + 1} + \sqrt{2}} = \frac{x^4-1}{(\sqrt[3]{x}-1)\cdot(\sqrt{x^4+1}+\sqrt{2})}$$
but I don't see if this leads anywhere.
 [/INST] HINT:
$$\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1} =\frac{x^4+1-2}{x-1}\cdot\frac{\sqrt[3]{x^2}+\sqrt[3]x+1}{\sqrt{x^4 + 1} + \sqrt{2}}$$
Now as $x\to1,x\ne1,x-1\ne0,$ so we can safely cancel out $x-1$ 
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