Question

The limit \( \lim_{x\to0} \frac{\sqrt{2+x} - \sqrt{2-x}}{x} \) is equal to:

• A. \( \frac{1}{\sqrt{2}} \)
• B. \( \sqrt{2} \)
• C. \( 0 \)
• D. Does not exist
• E. \( \frac{1}{2\sqrt{2}} \)

Hint:

For limits resulting in \( 0/0 \) with radicals, rationalization is the standard algebraic approach. If you are comfortable with derivatives, L'Hôpital's Rule is often faster: \( \text{Deriv}(\text{num}) / \text{Deriv}(\text{den}) = [\frac{1}{2\sqrt{2+x}} - \frac{-1}{2\sqrt{2-x}}] / 1 \), which yields the same result at \( x=0 \).

Answer 1

The Correct Option is A

Concept: Direct substitution of \( x = 0 \) results in an indeterminate form of \( \frac{0}{0} \). To resolve this for expressions involving square roots, we use the method of rationalization, which involves multiplying the numerator and denominator by the conjugate of the numerator. Alternatively, L'Hôpital's Rule can be applied.

Step 1: Rationalizing the numerator.
Multiply by the conjugate \( \sqrt{2+x} + \sqrt{2-x} \): \[ \lim_{x\to0} \frac{(\sqrt{2+x} - \sqrt{2-x})(\sqrt{2+x} + \sqrt{2-x})}{x(\sqrt{2+x} + \sqrt{2-x})} \] Applying the identity \( (a-b)(a+b) = a^2 - b^2 \): \[ = \lim_{x\to0} \frac{(2+x) - (2-x)}{x(\sqrt{2+x} + \sqrt{2-x})} \] \[ = \lim_{x\to0} \frac{2x}{x(\sqrt{2+x} + \sqrt{2-x})} \]

Step 2: Simplifying and evaluating the limit.
Cancel \( x \) from the numerator and denominator: \[ = \lim_{x\to0} \frac{2}{\sqrt{2+x} + \sqrt{2-x}} \] Now, substitute \( x = 0 \): \[ = \frac{2}{\sqrt{2+0} + \sqrt{2-0}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \]