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$$\begin{align*}&\lim_{x\to0}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}=\\ &\lim_{x\to0}\left(\frac{(1+x)-(1-x)}{(1+x)-(1-x)}\cdot\frac{\sqrt[3]{(1+x)^2}+\sqrt[3]{(1+x)(1-x)}+\sqrt[3]{(1-x)^2}}{\sqrt{1+x}+\sqrt{1-x}}\right) \end{align*}$$

My teacher was using this to calculate the value of the first limit as seen above.

I am not sure how he pull this huge rabbit out of the hat XD.

Also I am not sure why is it useful...

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up vote 1 down vote accepted

To get from the first line to the second line, your teacher is multiplying by $$ \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \cdot \frac{\sqrt[3]{(1+x)^2} + \sqrt[3]{(1+x)(1-x)} + \sqrt[3]{(1-x)^2}}{\sqrt[3]{(1+x)^2} + \sqrt[3]{(1+x)(1-x)} + \sqrt[3]{(1-x)^2}}. $$

Why did he make this choice? He is using the identities $$ (\sqrt{a} - \sqrt{b}) \cdot (\sqrt{a} + \sqrt{b}) = a - b $$ and $$ (\sqrt[3]{a} - \sqrt[3]{b}) \cdot (\sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}) = a - b. $$

As with many limit problems, the reason for these manipulations is to try to obtain an algebraic cancellation to deal with the fact that both the numerator and denominator are approaching zero.

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Your teacher has used two factorization identities: $a-b=\frac {a^2-b^2}{a+b}$ and $c-d=\frac{c^3-d^3}{c^2+cd+d^2}$. The first is represented by the numerator on the left, the numerator of the first term on the right, and the denominator of the second term on the right. The second is represented by the other three terms. As to why it is useful, note that the first fraction on the right is $1$ and the second is no longer going to $\frac 00$, so you can just evaluate it. This is an example of multiplying by the conjugate, often useful to clear indeterminate forms.

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You should use the following identities: $$a^2-b^2=(a-b)(a+b)$$ $$c^3-d^3=(c-d)(c^2+cd+d^2)$$ (prove by opening brackets) for $$a=\sqrt{1+x}, \hspace{5pt} b=\sqrt{1-x}, \hspace{5pt} c=\sqrt[3]{1+x}, \hspace{5pt} d=\sqrt[3]{1-x}$$ Then you multiply your function in the limit by $\frac{a+b}{c^2+cd+d^2}\cdot \frac{c^2+cd+d^2}{a+b}$, which leaves an equal expression. This is useful since instead of getting $\frac00$, you get an expression where you can substitute $x=0$.

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