# 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.

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

• Hm, this series adds up to 12, multiplied by $\frac{1}{\sqrt{x^4+1}+\sqrt{2}}$ , which is $\frac{12}{2\cdot\sqrt{2}}$ as x approaches 1, but the answer is $3\cdot\sqrt{2}$ Commented Jan 1, 2015 at 12:51
• @Innkeeper $\frac {12}{2\sqrt 2}\times \frac {\sqrt 2}{\sqrt 2}=\frac {12\sqrt 2}{4}=3\sqrt 2$ Commented Jan 1, 2015 at 13:10
• Where does the $\frac{\sqrt{2}}{\sqrt{2}}$ come from? Commented Jan 1, 2015 at 13:15
• @Innkeeper It is another way of writing $1$ and is chosen to eliminate the $\sqrt 2$ from the denominator. Commented Jan 1, 2015 at 13:16
• Oh, ok, $\frac{12}{2\sqrt{2}}$ is not at all $6\sqrt{2}$. I feel really stupid now, thanks for the help. Commented Jan 1, 2015 at 13:23

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 $x-1$

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

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

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

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