The value of `lim_(xrarr0)((1+6x)^((1)/(3))-(1+4x)^((1)/(2)))/(x^(2))` is equal to

Options

A. `1`
B. `2`
C. `-1`
D. `-2`

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{(1 + 6x)^{\frac{1}{3}} - (1 + 4x)^{\frac{1}{2}}}{x^2}, \]

we start by substituting \(x = 0\):

1. **Substituting \(x = 0\)**:
\[ \frac{(1 + 6 \cdot 0)^{\frac{1}{3}} - (1 + 4 \cdot 0)^{\frac{1}{2}}}{0^2} = \frac{1 - 1}{0} = \frac{0}{0}. \]
This gives us an indeterminate form \(0/0\), so we can apply L'Hôpital's Rule.

2. **Applying L'Hôpital's Rule**:
Differentiate the numerator and the denominator separately:
- The numerator is \( (1 + 6x)^{\frac{1}{3}} - (1 + 4x)^{\frac{1}{2}} \).
- The denominator is \( x^2 \).

Differentiate the numerator:
\[ \frac{d}{dx}[(1 + 6x)^{\frac{1}{3}}] = \frac{1}{3}(1 + 6x)^{-\frac{2}{3}} \cdot 6 = 2(1 + 6x)^{-\frac{2}{3}}, \]
\[ \frac{d}{dx}[(1 + 4x)^{\frac{1}{2}}] = \frac{1}{2}(1 + 4x)^{-\frac{1}{2}} \cdot 4 = 2(1 + 4x)^{-\frac{1}{2}}. \]

Therefore, the derivative of the numerator is:
\[ 2(1 + 6x)^{-\frac{2}{3}} - 2(1 + 4x)^{-\frac{1}{2}}. \]

The derivative of the denominator is:
\[ \frac{d}{dx}[x^2] = 2x. \]

Thus, we have:
\[ \lim_{x \to 0} \frac{2(1 + 6x)^{-\frac{2}{3}} - 2(1 + 4x)^{-\frac{1}{2}}}{2x} = \lim_{x \to 0} \frac{(1 + 6x)^{-\frac{2}{3}} - (1 + 4x)^{-\frac{1}{2}}}{x}. \]

3. **Substituting \(x = 0\) again**:
Again substituting \(x = 0\):
\[ \frac{(1 + 6 \cdot 0)^{-\frac{2}{3}} - (1 + 4 \cdot 0)^{-\frac{1}{2}}}{0} = \frac{1 - 1}{0} = \frac{0}{0}. \]
We apply L'Hôpital's Rule again.

4. **Differentiating again**:
Differentiate the new numerator:
\[ \frac{d}{dx}[(1 + 6x)^{-\frac{2}{3}}] = -\frac{2}{3}(1 + 6x)^{-\frac{5}{3}} \cdot 6 = -4(1 + 6x)^{-\frac{5}{3}}, \]
\[ \frac{d}{dx}[(1 + 4x)^{-\frac{1}{2}}] = -\frac{1}{2}(1 + 4x)^{-\frac{3}{2}} \cdot 4 = -2(1 + 4x)^{-\frac{3}{2}}. \]

Therefore, the new numerator is:
\[ -4(1 + 6x)^{-\frac{5}{3}} + 2(1 + 4x)^{-\frac{3}{2}}. \]

The derivative of the denominator \(x\) is \(1\).

Thus, we have:
\[ \lim_{x \to 0} \left(-4(1 + 6x)^{-\frac{5}{3}} + 2(1 + 4x)^{-\frac{3}{2}}\right). \]

5. **Substituting \(x = 0\) again**:
\[ -4(1 + 0)^{-\frac{5}{3}} + 2(1 + 0)^{-\frac{3}{2}} = -4 \cdot 1 + 2 \cdot 1 = -4 + 2 = -2. \]

Thus, the value of the limit is:

\[ \boxed{-2}. \]