Evaluate the limit: lim (x->1) (x^(1/6) - 1) / (x - 1)

Steps to Solve

1. Recognize the indeterminate form

   When we substitute $x = 1$ into the expression, we get $\frac{1^{1/6} - 1}{1 - 1} = \frac{0}{0}$, which is an indeterminate form. This means we can apply L'Hopital's Rule.

2. Apply L'Hopital's Rule

   L'Hopital's Rule states that if $\lim_{x\to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}$, provided the limit exists.

   In our case, $f(x) = x^{1/6} - 1$ and $g(x) = x - 1$. We need to find their derivatives.

3. Calculate the derivatives

   The derivative of $f(x)$ with respect to $x$ is: $f'(x) = \frac{d}{dx}(x^{1/6} - 1) = \frac{1}{6}x^{\frac{1}{6} - 1} = \frac{1}{6}x^{-5/6}$

   The derivative of $g(x)$ with respect to $x$ is: $g'(x) = \frac{d}{dx}(x - 1) = 1$

4. Evaluate the limit of the derivatives

   Now we evaluate the limit of the ratio of the derivatives: $\lim_{x\to 1} \frac{f'(x)}{g'(x)} = \lim_{x\to 1} \frac{\frac{1}{6}x^{-5/6}}{1} = \lim_{x\to 1} \frac{1}{6}x^{-5/6}$

   Substitute $x = 1$ into the expression: $\frac{1}{6}(1)^{-5/6} = \frac{1}{6}(1) = \frac{1}{6}$