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?
 A: Since you are allowed to replace $x$ with $y^6$,
$$ \lim_{x\to 1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\lim_{y\to 1}\frac{y^2-1}{y^3-1}=\lim_{y\to 1}\frac{y+1}{y^2+y+1}=\color{red}{\frac{2}{3}}.$$
A: Here's another variation of the theme:
Since we observe
$$\lim_{x\rightarrow 1}{\sqrt[3]{x}-1}=\lim_{x\rightarrow 1}{\sqrt{x}-1}=0$$
we can apply L’Hospital’s Rule

We obtain
  $$\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}
=\lim_{x\rightarrow 1}\frac{x^{\frac{1}{3}}-1}{x^{\frac{1}{2}}-1}
=\lim_{x\rightarrow 1}\frac{\frac{1}{3}x^{-\frac{2}{3}}}{\frac{1}{2}x^{-\frac{1}{2}}}
=\frac{\frac{1}{3}}{\frac{1}{2}}
=\frac{2}{3}
$$

Note: We apply L’Hospital’s Rule to $\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}$ in order to cope with the indefinite expression $\frac{0}{0}$. We could ask how the answer by @JackDAurizio deals with it. In fact this is done by cancelling $y-1$ in $\lim_{y\rightarrow 1}\frac{y^3-1}{y^2-1}$. We see this way that $1$ is a removable singularity.
A: Another way (Taylor expansions, again)*: write $x=1+h$, so that you are looking at the limit when $h\to 0$. Recall that, for $\alpha > 0$ the first-order Taylor expansion of $(1+h)^\alpha$ around $0$ is
$$
(1+h)^\alpha = \alpha h + o(h).
$$
Then, you can compute your limit as the limit around $0$ of
$$
\frac{(1+h)^{1/3}-1}{(1+h)^{1/2}-1} = \frac{1+\frac{1}{3}h+o(h)-1}{1+\frac{1}{2}h+o(h)-1} = \frac{\frac{1}{3}h+o(h)}{\frac{1}{2}h+o(h)} = \frac{2+o(1)}{3+o(1)} \xrightarrow[h\to0]{} \frac{2}{3}
$$
* A systematic, if not always most elegant, approach.
A: 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 $b=1$.
A: Multiply through by $\frac{\sqrt{x}+1}{\sqrt{x}+1}$ to get the equivalent limit  $$\lim_{x \to 1} \frac{x^{5/6}+x^{1/3}-x^{1/2}-1}{x-1}$$ Then use L'Hospital's rule and plug in $x=1$.
A: With conjugates: the rationale  behing conjugates is to use the factorisation  of $x^r-1$, so the conjgate expression of $\sqrt[3]{x}-1$ is $\,\sqrt[3]{x^2}+\sqrt[3]{x}+1$, so that :
$$\frac{\sqrt[3]{x}-1}{\sqrt x-1}=\frac{\sqrt x+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\xrightarrow[x\to 1]{}\frac23.$$
