`lim_(h to 0) [1/(h root(3)(8+h))-1/(2h)]`=

Options

A. `1/12`
B. `-4/3`
C. `-16/3`
D. `-1/48`

Answer

To solve the limit

\[ \lim_{h \to 0} \left[ \frac{1}{h \sqrt[3]{8+h}} - \frac{1}{2h} \right], \]

we will follow these steps:

### Step 1: Combine the fractions
First, we need to combine the two fractions under a common denominator. The common denominator will be \(2h \sqrt[3]{8+h}\):

\[ \frac{1}{h \sqrt[3]{8+h}} - \frac{1}{2h} = \frac{2 - \sqrt[3]{8+h}}{2h \sqrt[3]{8+h}}. \]

### Step 2: Rewrite the limit
Now we can rewrite the limit as:

\[ \lim_{h \to 0} \frac{2 - \sqrt[3]{8+h}}{2h \sqrt[3]{8+h}}. \]

### Step 3: Substitute \(h = 0\)
If we substitute \(h = 0\) directly, we get:

\[ \frac{2 - \sqrt[3]{8}}{0} = \frac{2 - 2}{0} = \frac{0}{0}, \]

which is an indeterminate form. Therefore, we can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:

- The derivative of the numerator \(2 - \sqrt[3]{8+h}\) is:

\[ -\frac{1}{3}(8+h)^{-\frac{2}{3}} \cdot 1 = -\frac{1}{3(8+h)^{\frac{2}{3}}}. \]

- The derivative of the denominator \(2h \sqrt[3]{8+h}\) is:

Using the product rule:

\[ 2 \sqrt[3]{8+h} + 2h \cdot \frac{1}{3}(8+h)^{-\frac{2}{3}} \cdot 1 = 2 \sqrt[3]{8+h} + \frac{2h}{3(8+h)^{\frac{2}{3}}}. \]

### Step 5: Rewrite the limit again
Now we can rewrite the limit as:

\[ \lim_{h \to 0} \frac{-\frac{1}{3(8+h)^{\frac{2}{3}}}}{2 \sqrt[3]{8+h} + \frac{2h}{3(8+h)^{\frac{2}{3}}}}. \]

### Step 6: Substitute \(h = 0\) again
Now substituting \(h = 0\):

- The numerator becomes:

\[ -\frac{1}{3(8)^{\frac{2}{3}}} = -\frac{1}{3 \cdot 4} = -\frac{1}{12}. \]

- The denominator becomes:

\[ 2 \cdot 2 + 0 = 4. \]

### Step 7: Final calculation
Thus, we have:

\[ \lim_{h \to 0} \frac{-\frac{1}{12}}{4} = -\frac{1}{48}. \]

### Final Answer
The limit is:

\[ \boxed{-\frac{1}{48}}. \]