Let `L=lim_(xrarr0)(b-sqrt(b^(2)+x^(2))-(x^(2))/(4))/(x^(4)), b gt 0`. If L is finite, then

Options

A. `b=2`
B. `b=1`
C. `b=(1)/(3)`
D. None of these

Answer

To solve the limit problem given by \( L = \lim_{x \to 0} \frac{b - \sqrt{b^2 + x^2} - \frac{x^2}{4}}{x^4} \) where \( b > 0 \), we will follow these steps:

### Step 1: Analyze the limit
As \( x \to 0 \), both the numerator and denominator approach 0. Thus, we have a \( \frac{0}{0} \) indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator separately.

The numerator is:
\[ f(x) = b - \sqrt{b^2 + x^2} - \frac{x^2}{4} \]
The denominator is:
\[ g(x) = x^4 \]

Taking the derivatives:
1. Derivative of the numerator \( f'(x) \):
\[ f'(x) = 0 - \frac{1}{2\sqrt{b^2 + x^2}} \cdot 2x - \frac{1}{2}x = -\frac{x}{\sqrt{b^2 + x^2}} - \frac{x}{2} \]
Simplifying gives:
\[ f'(x) = -x \left( \frac{1}{\sqrt{b^2 + x^2}} + \frac{1}{2} \right) \]

2. Derivative of the denominator \( g'(x) \):
\[ g'(x) = 4x^3 \]

### Step 3: Rewrite the limit using L'Hôpital's Rule
Now we can rewrite the limit:
\[ L = \lim_{x \to 0} \frac{-x \left( \frac{1}{\sqrt{b^2 + x^2}} + \frac{1}{2} \right)}{4x^3} \]
This simplifies to:
\[ L = \lim_{x \to 0} \frac{-\left( \frac{1}{\sqrt{b^2 + x^2}} + \frac{1}{2} \right)}{4x^2} \]

### Step 4: Evaluate the limit as \( x \to 0 \)
Substituting \( x = 0 \):
\[ L = \frac{-\left( \frac{1}{\sqrt{b^2}} + \frac{1}{2} \right)}{0} \]
This indicates that the limit will be infinite unless the numerator equals zero.

### Step 5: Set the numerator to zero
For the limit \( L \) to be finite, we need:
\[ -\left( \frac{1}{\sqrt{b^2}} + \frac{1}{2} \right) = 0 \]
This leads to:
\[ \frac{1}{\sqrt{b^2}} + \frac{1}{2} = 0 \]
Since \( b > 0 \), we can solve:
\[ \frac{1}{b} + \frac{1}{2} = 0 \implies \frac{1}{b} = -\frac{1}{2} \]
This is not possible since \( b \) must be positive.

### Conclusion
Thus, for \( L \) to be finite, we conclude that:
\[ b = -2 \]
However, since \( b > 0 \) is given, there is no valid solution for \( b \) under the constraints provided. Therefore, the answer is that there is no valid \( b \) such that \( L \) is finite.