`lim_(x rarr 1)(x^(1//3) - 1)/(x^(1//6)-1)` is equal to

Options

A. 1
B. 4
C. 3
D. 2

Answer

To solve the limit \( \lim_{x \to 1} \frac{x^{1/3} - 1}{x^{1/6} - 1} \), we can follow these steps:

### Step 1: Substitute the limit
First, we will substitute \( x = 1 \) into the expression:

\[ \frac{1^{1/3} - 1}{1^{1/6} - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0} \]

This results in an indeterminate form \( \frac{0}{0} \), so we need to apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:

- The numerator \( x^{1/3} - 1 \) differentiates to:
\[ \frac{d}{dx}(x^{1/3}) = \frac{1}{3}x^{-2/3} \]

- The denominator \( x^{1/6} - 1 \) differentiates to:
\[ \frac{d}{dx}(x^{1/6}) = \frac{1}{6}x^{-5/6} \]

### Step 3: Rewrite the limit
Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 1} \frac{\frac{1}{3} x^{-2/3}}{\frac{1}{6} x^{-5/6}} \]

### Step 4: Simplify the expression
This simplifies to:

\[ \lim_{x \to 1} \frac{\frac{1}{3}}{\frac{1}{6}} \cdot \frac{x^{-2/3}}{x^{-5/6}} = \lim_{x \to 1} \frac{1}{3} \cdot \frac{6}{1} \cdot x^{(5/6 - 2/3)} \]

Now, we need to simplify the exponent:

\[ 5/6 - 2/3 = 5/6 - 4/6 = 1/6 \]

So we have:

\[ \lim_{x \to 1} \frac{6}{3} \cdot x^{1/6} = \lim_{x \to 1} 2 \cdot x^{1/6} \]

### Step 5: Evaluate the limit
Now substituting \( x = 1 \):

\[ 2 \cdot 1^{1/6} = 2 \]

Thus, the limit is:

\[ \boxed{2} \]