Limit $ \lim_{x \to 1} \frac{x^{\frac{1}{3}}-1}{\sqrt{x}-1}$ $ \lim_{x \to 1} \frac{x^{\frac{1}{3}}-1}{\sqrt{x}-1}$
How would I solve this limit question? Thanks in advance.
 A: Try change of variable $x = t^6$.
A: Setting $f(x)=x^{\frac{1}{3}}$ and $g(x)=\sqrt{x}$, we can rewrite the limit as 
$$ \lim_{x\to 1}\frac{x^{\frac{1}{3}}-1}{x-1}\lim_{x\to 1}\frac{x-1}{\sqrt{x}-1}=\frac{f^{\prime}(1)}{g^{\prime}(1)}=\frac{2}{3}$$
A: 
METHODOLOGY $1$:  L'Hospital's Rule

L'Hospital's rule works efficiently here.  We have
$$\begin{align}
\lim_{x\to 1}\left(\frac{x^{1/3}-1}{x^{1/2}-1}\right)&=\lim_{x\to 1}\left(\frac{\frac13x^{-2/3}}{\frac12 x^{-1/2}}\right)\\\\
&=\frac23\end{align}$$


METHODOLOGY $2$:  Non-Calculus Based

Alternatively, for those unfamiliar with differential calculus, we can simply write
$$\begin{align}
\lim_{x\to 1}\left(\frac{x^{1/3}-1}{x^{1/2}-1}\right)&=\lim_{x\to 1}\left(\frac{\frac{x^{1/3}-1}{x-1}}{\frac{x^{1/2}-1}{x-1}}\right)\\\\
&=\lim_{x\to 1}\left(\frac{x^{1/2}+1}{x^{2/3}+x^{1/3}+1}\right)\\\\
&=\frac23
\end{align}$$
A: Another possible way to do it.
Define $x=1+y$ which makes $$A=\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\frac{\sqrt[3]{1+y}-1}{\sqrt{1+y}-1}$$ and use the generalized binomial theorem or Taylor series which write $$(1+y)^a=1+a y+\frac{1}{2} a(a-1)  y^2+O\left(y^3\right)$$ which then gives $$A=\frac{1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)-1 } {1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)-1 }=\frac{\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right) }{\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right) }=\frac{\frac{1}{3}-\frac{y}{9}+O\left(y^2\right)} {\frac{1}{2}-\frac{y}{8}+O\left(y^2\right) }$$ Now, using long division $$A=\frac{2}{3}-\frac{y}{18}+O\left(y^2\right)$$ which shows the limit and how it is approached.
Making the problem more general, consider $$B=\frac{\sqrt[p]{x}-1}{\sqrt[q]{x}-1}$$ and the same procedure would give $$B=\frac{q}{p}+\frac{ (q-p)}{2 p^2}y+O\left(y^2\right)$$
A: Let $x=t^2$. Then this is the definition of the derivative of $t^{\frac 23}$ at $1$
