How do I simplify and evaluate the limit of $(\sqrt x - 1)/(\sqrt[3] x - 1)$ as $x\to 1$? Consider this limit:
$$ \lim_{x \to 1} \frac{\sqrt x -  1}{ \sqrt[3] x -  1} 
$$ 
The answer is given to be 2 in the textbook. 
Our math professor skipped this question telling us it is not in our syllabus, but how can it be solved?
 A: Note the following identities,
$$ y^2-1 = (y-1)(y+1) $$
$$ y^3-1 = (y-1)(y^2+y+1) $$
we can use these to rewrite the numerator and the denomiantor by substituting $\sqrt{x}$ and $\sqrt[3]{x}$ for $y$ respectively,
$$ x - 1 = (\sqrt{x}-1)(\sqrt{x}+1) \Rightarrow \sqrt{x}-1 = \frac{x-1}{\sqrt{x}+1} $$
$$ x - 1 = (\sqrt[3]{x}-1)(\sqrt[3]{x}^2+\sqrt[3]{x}+1) \Rightarrow \sqrt[3]{x}-1 = \frac{x-1}{\sqrt[3]{x}^2+\sqrt[3]{x}+1} $$
substituting this into the ratio you're taking the limit of we get,
$$ \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{\sqrt[3]{x}-1} = \lim_{x\rightarrow 1}  \frac{x-1}{\sqrt{x}+1} \frac{\sqrt[3]{x}^2+\sqrt[3]{x}+1} {x-1} = \frac32 $$
A: Using L'Hopital, $$\lim_{x\rightarrow 1} \frac{x^{1/2}-1}{x^{1/3}{-1}} = \lim_{x\rightarrow 1} \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{3x^{2/3}}} = \frac{3}{2}$$
A: The limit is solved as follows:
$$\lim_{x \to 1} \frac{\sqrt x -  1}{ \sqrt[3] x -  1} $$
$$=\lim_{x \to 1} \frac{\frac{\sqrt x -  1}{x-1}}{ \frac{\sqrt[3] x -  1}{x-1}} $$
$$=\frac{\lim_{x \to 1}  \frac{\sqrt x -  1}{x-1}}{ \lim_{x \to 1}  \frac {\sqrt[3] x -  1}{x-1}} $$
$$=\frac{\frac{1}{2}\cdot 1^{-\frac{1}{2}}}{\frac{1}{3}\cdot 1^{-\frac{2}{3}}} $$
$$=\frac{3}{2}$$
using the common limit formula : $$\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$$
A: Notice, $$\lim_{x\to 1}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}$$
$$=\lim_{x\to 1}\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{(\sqrt[3]{x}-1)(x^{2/3}+1+\sqrt [3]x)}\cdot \frac{(x^{2/3}+1+\sqrt [3]x)}{(\sqrt x+1)}$$
$$=\lim_{x\to 1}\frac{(x-1)}{(x-1)}\cdot \frac{(x^{2/3}+1+\sqrt[3] x)}{(\sqrt x+1)}$$
$$=\lim_{x\to 1}\frac{(x^{2/3}+1+\sqrt[3] x)}{(\sqrt x+1)}$$
$$=\frac{1+1+1}{1+1}=\color{red}{\frac 32}$$
A: As lcm$(2,3)=6$
let $\sqrt[6]x=y\implies\sqrt[3]x=y^2, \sqrt x=y^3$
$$\lim_{x\to1}\dfrac{\sqrt x-1}{\sqrt[3]x-1}=\lim_{y\to1}\dfrac{y^3-1}{y^2-1}=\lim_{y\to1}\dfrac{(y-1)(y^2+y+1)}{(y-1)(y+1)}$$
Safely cancel out $y-1$ as $y\to1, y-1\to0\implies y-1\ne0$
OR
set $\sqrt[6]x=y+1$ to get $$\lim_{x\to1}\dfrac{\sqrt x-1}{\sqrt[3]x-1}=\lim_{y\to0}\dfrac{(1+y)^3-1}{(1+y)^2-1}=\lim_{y\to0}\dfrac{3y+3y^2+y^3}{2y+y^2}=\cdots$$
A: Another way : change variable $x=1+y$; so $$A=\frac{\sqrt x -  1}{ \sqrt[3] x -  1}=\frac{\sqrt{1+y} -  1}{ \sqrt[3] {1+y} -  1}$$ Now, using the fact that, close to $y=0$ (using the generalized binomial theorem as lulu commented)  $$(1+y)^a=1+a y+\frac{1}{2}a \left(a-1\right) y^2+O\left(y^3\right)$$  which makes $$A=\frac{1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)-1}{1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)-1}\approx \frac{\frac{y}{2}-\frac{y^2}{8}}{\frac{y}{3}-\frac{y^2}{9}}=\frac{\frac{1}{2}-\frac{y}{8}}{\frac{1}{3}-\frac{y}{9}}$$ Now make $y\to0$ to get the result.
You can even get more if you know long division. Omitting the high order terms, the last expression is $\sim\frac{3}{2}+\frac{y}{8}$ which reveals not only the limit but also how it is approached.
A: Here's a similar approach to Spencer and Harish, but I use a substitution to make it a little easier to read.
First, we eliminate the fractional exponents by substituting $x=u^6$.
Note that $\lim_{x \to 1} u = x = 1$
$$\begin{align}\\
& \lim_{x \to 1} \frac{\sqrt x -  1}{ \sqrt[3] x -  1}\\
 = & \lim_{u \to 1} \frac{u^3 - 1}{u^2 - 1}\\
 = & \lim_{u \to 1} \frac{(u - 1)(u^2+u+1)}{(u-1)(u+1)}\\
 = & \lim_{u \to 1} \frac{u^2+u+1}{u+1}\\
 = & \frac{3}{2}
\end{align}$$
