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?
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 $\lim_{x \to 1} u = x = 1$
$$\begin{align}\\ & \lim_{x \to 1} \frac{\sqrt x - 1}{ \sqrt[3] x - 1}\\ = & \lim_{u \to 1} \frac{u^3 - 1}{u^2 - 1}\\ = & \lim_{u \to 1} \frac{(u - 1)(u^2+u+1)}{(u-1)(u+1)}\\ = & \lim_{u \to 1} \frac{u^2+u+1}{u+1}\\ = & \frac{3}{2} \end{align}$$
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,
$$ \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{\sqrt[3]{x}-1} = \lim_{x\rightarrow 1} \frac{x-1}{\sqrt{x}+1} \frac{\sqrt[3]{x}^2+\sqrt[3]{x}+1} {x-1} = \frac32 $$
Using L'Hopital, $$\lim_{x\rightarrow 1} \frac{x^{1/2}-1}{x^{1/3}{-1}} = \lim_{x\rightarrow 1} \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{3x^{2/3}}} = \frac{3}{2}$$
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 : $$\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$$