Let `kgt0 and lambda =lim_(xrarr0) (k(1-4sqrt(k^2-x^2)))/(x^2sqrt(k^2-x^2))` be finite. Then the value of `lambdak`, is

Options

A. `lambda=8,k=(1)/(2)`
B. `lambda=8,k=(1)/(4)`
C. `lambda=4, k=(1)/(2)`
D. `6lambda=4,k=(1)/(4)`

Explanation

To solve the problem, we need to evaluate the limit:

\[ \lambda = \lim_{x \to 0} \frac{k(1 - 4\sqrt{k^2 - x^2})}{x^2 \sqrt{k^2 - x^2}} \]

where \( k > 0 \) and find the value of \( \lambda k \).

### Step-by-Step Solution:

**Step 1: Analyze the limit.**

As \( x \to 0 \), \( \sqrt{k^2 - x^2} \) approaches \( k \). Thus, we can rewrite the limit:

\[ \lambda = \lim_{x \to 0} \frac{k(1 - 4\sqrt{k^2 - x^2})}{x^2 \sqrt{k^2 - x^2}} = \lim_{x \to 0} \frac{k(1 - 4k)}{x^2 k} = \lim_{x \to 0} \frac{1 - 4k}{x^2} \]

**Step 2: Ensure the limit is finite.**

For \( \lambda \) to be finite, the numerator must equal zero when \( x \to 0 \). Therefore, we set:

\[ 1 - 4k = 0 \implies k = \frac{1}{4} \]

**Step 3: Substitute \( k \) back into the limit.**

Now substitute \( k = \frac{1}{4} \) back into the limit to find \( \lambda \):

\[ \lambda = \lim_{x \to 0} \frac{\frac{1}{4}(1 - 4\sqrt{\left(\frac{1}{4}\right)^2 - x^2})}{x^2 \sqrt{\left(\frac{1}{4}\right)^2 - x^2}} \]

Calculating \( \left(\frac{1}{4}\right)^2 = \frac{1}{16} \):

\[ \lambda = \lim_{x \to 0} \frac{\frac{1}{4}(1 - 4\sqrt{\frac{1}{16} - x^2})}{x^2 \sqrt{\frac{1}{16} - x^2}} \]

**Step 4: Simplify the limit.**

Now simplify the expression:

\[ \lambda = \lim_{x \to 0} \frac{\frac{1}{4}(1 - 4\sqrt{\frac{1}{16} - x^2})}{x^2 \sqrt{\frac{1}{16} - x^2}} \]

Rationalizing the numerator:

\[ 1 - 4\sqrt{\frac{1}{16} - x^2} = 1 - 4\sqrt{\frac{1}{16}(1 - 16x^2)} = 1 - \frac{4}{4}\sqrt{1 - 16x^2} = 1 - \sqrt{1 - 16x^2} \]

**Step 5: Apply the limit.**

Now we can apply the limit:

\[ \lambda = \lim_{x \to 0} \frac{(1 - \sqrt{1 - 16x^2})}{x^2 \cdot \frac{1}{4}\sqrt{1 - 16x^2}} = \lim_{x \to 0} \frac{16x^2}{x^2 \cdot \frac{1}{4}\sqrt{1 - 16x^2}} = \lim_{x \to 0} \frac{16}{\frac{1}{4}\sqrt{1 - 16x^2}} = 64 \]

**Step 6: Calculate \( \lambda k \).**

Now we find \( \lambda k \):

\[ \lambda k = 64 \cdot \frac{1}{4} = 16 \]

### Final Answer:

\[ \lambda k = 16 \]