Simple limit problem without L'Hospital's rule

Question

$$\lim_{x \to 1}\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1}$$

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.

Answer 1

Note that if you put $y=\sqrt[3] x$ then $$\frac {x^4-1}{\sqrt[3]x-1}=\frac {y^{12}-1}{y-1}= y^{11}+y^{10}+\dots +1$$

Answer 2

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 $x-1$

Answer 3

Let $\sqrt[3]x-1=y\implies x=(1+y)^3,x^4=[(1+y)^3]^4=(1+y)^{12}$

$$\lim_{x \to 1}\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1}=\lim_{y\to0}\frac{\sqrt{(1+y)^{12}+1}-\sqrt2}y$$

$$=\lim_{y\to0}\frac{(1+y)^{12}+1-2}{(\sqrt{(1+y)^{12}+1}+\sqrt2)y}$$

$$=\lim_{y\to0}\frac{1+1+12y+O(y^2)-2}y\cdot\frac1{\lim_{y\to0}(\sqrt{(1+y)^{12}+1}+\sqrt2)}$$