Question

If \( f(x) = \begin{cases \frac{\sqrt{a^2 - ax - x^2} - \sqrt{x^2 + ax + a^2}}{\sqrt{a + x} - \sqrt{a - x}}, & x \ne 0
K, & x = 0 \end{cases} \) is continuous at \( x = 0 \), then \( K = \)}

• A. \( -\sqrt{a} \)
• B. \( \sqrt{a} \)
• C. \( -1 \)
• D. \( a + \sqrt{a} \)

Hint:

To evaluate limits of the form \(\frac{0}{0}\), we can use L'Hôpital's rule or rationalize the numerator or denominator.

Answer 1

The Correct Option is A

Since \(f(x)\) is continuous at \(x = 0\), we must have \[\lim_{x \to 0} f(x) = f(0) = K.\]Thus, we need to find \[\lim_{x \to 0} \frac{\sqrt{a^2
- ax
- x^2}
- \sqrt{x^2 + ax + a^2}}{\sqrt{a + x}
- \sqrt{a
- x}}.\]We have \begin{align*} &\lim_{x \to 0} \frac{\sqrt{a^2
- ax
- x^2}
- \sqrt{x^2 + ax + a^2}}{\sqrt{a + x}
- \sqrt{a
- x}}
&= \lim_{x \to 0} \frac{\sqrt{a^2
- ax
- x^2}
- \sqrt{x^2 + ax + a^2}}{\sqrt{a + x}
- \sqrt{a
- x}} \cdot \frac{\sqrt{a + x} + \sqrt{a
- x}}{\sqrt{a + x} + \sqrt{a
- x}}
&= \lim_{x \to 0} \frac{\sqrt{a^2
- ax
- x^2}
- \sqrt{x^2 + ax + a^2}}{2x} \cdot (\sqrt{a + x} + \sqrt{a
- x})
&= \lim_{x \to 0} \frac{\sqrt{a^2
- ax
- x^2}
- \sqrt{x^2 + ax + a^2}}{2x} \cdot 2 \sqrt{a}
&= \sqrt{a} \lim_{x \to 0} \frac{\sqrt{a^2
- ax
- x^2}
- \sqrt{x^2 + ax + a^2}}{x}
&= \sqrt{a} \lim_{x \to 0} \frac{\sqrt{a^2
- ax
- x^2}
- \sqrt{x^2 + ax + a^2}}{x} \cdot \frac{\sqrt{a^2
- ax
- x^2} + \sqrt{x^2 + ax + a^2}}{\sqrt{a^2
- ax
- x^2} + \sqrt{x^2 + ax + a^2}}
&= \sqrt{a} \lim_{x \to 0} \frac{(a^2
- ax
- x^2)
- (x^2 + ax + a^2)}{x (\sqrt{a^2
- ax
- x^2} + \sqrt{x^2 + ax + a^2})}
&= \sqrt{a} \lim_{x \to 0} \frac{
-2ax
- 2x^2}{x (\sqrt{a^2
- ax
- x^2} + \sqrt{x^2 + ax + a^2})}
&= \sqrt{a} \lim_{x \to 0} \frac{
-2a
- 2x}{\sqrt{a^2
- ax
- x^2} + \sqrt{x^2 + ax + a^2}}
&= \sqrt{a} \cdot \frac{
-2a}{2 \sqrt{a^2}}
&= \sqrt{a} \cdot \frac{
-2a}{2a}
&=
-\sqrt{a}. \end{align*}Therefore, \(K =
-\sqrt{a}\).