The value of `lim_(xrarr 4) frac{(5 + x)^(1/3) - (1 + 2x)^(1/3)}{(5 + x)^(1/2) - (1+ 2x)^(1/2)}` is

Options

A. `2(9^(1/3))`
B. `2/(9^(1/3))`
C. `2(9^(1/3))/9`
D. `2/(3^(1/3))`

Answer

To solve the limit problem given by

\[ \lim_{x \to 4} \frac{(5 + x)^{1/3} - (1 + 2x)^{1/3}}{(5 + x)^{1/2} - (1 + 2x)^{1/2}}, \]

we will follow these steps:

### Step 1: Direct Substitution
First, we substitute \(x = 4\) into the expression to check if we can directly evaluate the limit.

\[ (5 + 4)^{1/3} - (1 + 2 \cdot 4)^{1/3} = 9^{1/3} - 9^{1/3} = 0, \]
\[ (5 + 4)^{1/2} - (1 + 2 \cdot 4)^{1/2} = 9^{1/2} - 9^{1/2} = 0. \]

This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

if the limit on the right exists.

### Step 3: Differentiate the Numerator and Denominator
Let \(f(x) = (5 + x)^{1/3} - (1 + 2x)^{1/3}\) and \(g(x) = (5 + x)^{1/2} - (1 + 2x)^{1/2}\).

#### Differentiate \(f(x)\):
Using the chain rule,
\[ f'(x) = \frac{1}{3}(5 + x)^{-2/3} \cdot 1 - \frac{1}{3}(1 + 2x)^{-2/3} \cdot 2 = \frac{1}{3}(5 + x)^{-2/3} - \frac{2}{3}(1 + 2x)^{-2/3}. \]

#### Differentiate \(g(x)\):
Similarly,
\[ g'(x) = \frac{1}{2}(5 + x)^{-1/2} \cdot 1 - \frac{1}{2}(1 + 2x)^{-1/2} \cdot 2 = \frac{1}{2}(5 + x)^{-1/2} - (1 + 2x)^{-1/2}. \]

### Step 4: Evaluate the Limit Again
Now we can evaluate the limit:

\[ \lim_{x \to 4} \frac{f'(x)}{g'(x)} = \lim_{x \to 4} \frac{\frac{1}{3}(5 + x)^{-2/3} - \frac{2}{3}(1 + 2x)^{-2/3}}{\frac{1}{2}(5 + x)^{-1/2} - (1 + 2x)^{-1/2}}. \]

Substituting \(x = 4\):

\[ f'(4) = \frac{1}{3}(9)^{-2/3} - \frac{2}{3}(9)^{-2/3} = \frac{1}{3} \cdot \frac{1}{9^{2/3}} - \frac{2}{3} \cdot \frac{1}{9^{2/3}} = -\frac{1}{3} \cdot \frac{1}{9^{2/3}}. \]

\[ g'(4) = \frac{1}{2}(9)^{-1/2} - (9)^{-1/2} = \frac{1}{2} \cdot \frac{1}{9^{1/2}} - \frac{1}{9^{1/2}} = -\frac{1}{2} \cdot \frac{1}{9^{1/2}}. \]

### Step 5: Final Calculation
Now substituting these into the limit:

\[ \lim_{x \to 4} \frac{-\frac{1}{3} \cdot \frac{1}{9^{2/3}}}{-\frac{1}{2} \cdot \frac{1}{9^{1/2}}} = \frac{\frac{1}{3} \cdot \frac{1}{9^{2/3}}}{\frac{1}{2} \cdot \frac{1}{9^{1/2}}} = \frac{2}{3} \cdot \frac{9^{1/2}}{9^{2/3}} = \frac{2}{3} \cdot \frac{1}{9^{1/6}}. \]

### Step 6: Simplify
This simplifies to:

\[ \frac{2}{3 \cdot 9^{1/6}} = \frac{2 \cdot 9^{1/3}}{9} = \frac{2}{3 \cdot 3^{2/3}} = \frac{2 \cdot 3^{1/3}}{9}. \]

Thus, the final answer is:

\[ \frac{2 \cdot 3^{1/3}}{9}. \]