<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] Let $\sqrt[3]x-1=y\implies x=(1+y)^3,x^4=[(1+y)^3]^4=(1+y)^{12}$
$$\lim_{x \to 1}\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1}=\lim_{y\to0}\frac{\sqrt{(1+y)^{12}+1}-\sqrt2}y$$
$$=\lim_{y\to0}\frac{(1+y)^{12}+1-2}{(\sqrt{(1+y)^{12}+1}+\sqrt2)y}$$
$$=\lim_{y\to0}\frac{1+1+12y+O(y^2)-2}y\cdot\frac1{\lim_{y\to0}(\sqrt{(1+y)^{12}+1}+\sqrt2)}$$
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