Limits and cube roots I'm rather stumped at the moment. I can graph the following equation but I'm having trouble solving it algebraically.
$$
\lim_{x\to 1} \frac{\sqrt[3]{x}-1}{\sqrt{x}-1}
$$
Where do I start?

This solution uses what is called: The conjugate expression. Using
\begin{equation*}
(a-b)(a+b)=a^{2}-b^{2}
\end{equation*}
with $a=\sqrt{x}$ and $b=1,$ you get
\begin{equation*}
\left( \sqrt{x}-1\right) (\sqrt{x}+1)=x-1.
\end{equation*}
[The conjugate of $(\sqrt{x}-1)$ is $(\sqrt{x}+1)$)]. The same way, using 
\begin{equation*}
(a-b)(a^{2}+ab+b^{2})=a^{3}-b^{3}
\end{equation*}
with $a=\sqrt[3]{x}$ and $b=1,$ you get 
\begin{equation*}
(\sqrt[3]{x}-1)(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)=x-1.
\end{equation*}
[The conjugate of $(\sqrt[3]{x}-1)$ is $(\left( \sqrt[3]{x}\right) ^{2}+\sqrt%
[3]{x}+1)$]. Therefore
\begin{eqnarray*}
\frac{(\sqrt[3]{x}-1)}{\left( \sqrt{x}-1\right) } &=&\frac{(\sqrt[3]{x}-1)}{%
\left( \sqrt{x}-1\right) }\frac{(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}%
+1)}{(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)}\frac{(\sqrt{x}+1)}{(%
\sqrt{x}+1)} \\
&=&\frac{(\sqrt[3]{x}-1)\cdot (\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)%
}{\left( \sqrt{x}-1\right) \cdot (\sqrt{x}+1)}\cdot \frac{(\sqrt{x}+1)}{%
(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)} \\
&=&\frac{x-1}{x-1}\cdot \frac{(\sqrt{x}+1)}{(\left( \sqrt[3]{x}\right) ^{2}+%
\sqrt[3]{x}+1)} \\
&=&\frac{(\sqrt{x}+1)}{(\left( \sqrt[3]{x}\right) ^{2}+\sqrt[3]{x}+1)}
\end{eqnarray*}
and the limit follows
\begin{equation*}
\lim_{x\rightarrow 1}\frac{(\sqrt[3]{x}-1)}{\left( \sqrt{x}-1\right) }%
=\lim_{x\rightarrow 1}\frac{(\sqrt{x}+1)}{(\left( \sqrt[3]{x}\right) ^{2}+%
\sqrt[3]{x}+1)}=\frac{(\sqrt{1}+1)}{(\left( \sqrt[3]{1}\right) ^{2}+\sqrt[3]{%
1}+1)}=\frac{2}{3}.
\end{equation*}
${\bf NOTE:}$ You can take it as a Definition for further uses: The conjugate
expression of $A$ (which may contains radicals) is the expression $B$ (which
may contains radicals), if $AB$ contains no radicals! For example, the
conjugate of $\sqrt[4]{x}-1$ is obtained from the formula 
\begin{equation*}
(a-b)(a^{3}+a^{2}b+ab^{2}+b^{3})=a^{4}-b^{4}
\end{equation*}
by taking $a=\sqrt[4]{x}$ and $b=1$.
