Calculate the lim $\lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1}$ $$
\lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1}
$$
I multiply it with:
$$
\frac{ \sqrt{x} + 1}{\sqrt{x} + 1}
$$
And I get :
$$
\lim_{x\to 1} \frac{ \sqrt{x}^2 - 1^2}{(\sqrt[3]{x} - 1) * (\sqrt{x} + 1)}
$$
But the solution is still division by 0 and not possible, so I think I've made a mistake somewhere..
 A: You basically want to insert $1$ written as a “conjugate” that removes the radical divided by itself. For $\sqrt{x}-1$ it is $\sqrt{x}+1$, for $\sqrt[3]{x}-1$ it is $\sqrt[3]{x^2}+\sqrt[3]{x}+1$. So you get
\begin{align}
\lim_{x\to1}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}&=
\lim_{x\to1}\left((\sqrt{x}-1)\frac{\sqrt{x}+1}{\sqrt{x}+1}\right)
  \left(\frac{1}{\sqrt[3]{x}-1}
    \frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\right)
\\[2ex]
&=\frac{x-1}{\sqrt{x}+1}\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{x-1}
\end{align}
and the indetermination goes away.
It's the same as Bill Dubuque's answer, actually, but without substitution.
A: Here is a version computationally close to L'Hospital's Rule, though it does not mention him. Note that
$$\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}= \frac{\frac{\sqrt{x}-1}{x-1} }{\frac{\sqrt[3]{x}-1}{x-1}   }.\tag{1}$$ 
If we look at the expression on the right of (1), we see that the limit as $x\to 1$ of the top is by definition the derivative of $\sqrt{x}$ at $x=1$. Similarly, the limit of the bottom is the derivative of $\sqrt[3]{x}$ at $x=1$. Compute these derivatives, using the ordinary "Power Rule."
A: Hint: $$\frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1} =\frac{3}{2} \cdot \frac{ e^\frac{\ln x}{2} - 1}{\frac{\ln x}{2}} \cdot \left( \frac{ e^\frac{\ln x}{3} - 1}{\frac{\ln x}{3}}\right)^{-1} $$
A: Hint $\ $ Let $\ X = x^{1/6}.\ $ Then it is $\ \dfrac{X^3-1}{X^2-1} = \dfrac{X^2+X+1}{X+1},\ $ no longer indeterminate.
A: Hint: You are looking to apply L'Hopital's Rule.
 This rule basically states the following: 
$$\lim_{x\to 1} \frac{f(x)}{g(x)} = \lim_{x\to 1} \frac{f'(x)}{g'(x)}$$

Note: you can keep applying this rule multiple times, so it can go on and on until you get the value of the limit.

So keep calculating the limit of the derivative until you get one that is not division by zero or undefined, that value is the answer is the limit of your original question.
