Question

Evaluate: $$ \lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x} $$

• A. 1
• B. 0
• C. \(\infty\)
• D. 0.5

Hint:

If limit results in \(\frac{0}{0}\), try rationalization or binomial expansion near \(x = 0\).

Answer 1

The Correct Option is A

To evaluate the limit \( \lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x} \), we begin by simplifying the expression using rationalization. The conjugate of \( \sqrt{1 + x} - \sqrt{1 - x} \) is \( \sqrt{1 + x} + \sqrt{1 - x} \).

Multiply the numerator and the denominator by the conjugate:

$$\frac{\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(\sqrt{1+x}+\sqrt{1-x}\right)}{x\left(\sqrt{1+x}+\sqrt{1-x}\right)}$$

This simplifies the numerator using the difference of squares:

$$=\frac{(1+x)-(1-x)}{x(\sqrt{1+x}+\sqrt{1-x})}$$

$$=\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$$

We can cancel \( x \) in the numerator and denominator (assuming \( x\neq0 \)):

$$=\frac{2}{\sqrt{1+x}+\sqrt{1-x}}$$

Now compute the limit as \( x \to 0 \):

$$\lim_{x \to 0} \frac{2}{\sqrt{1+x}+\sqrt{1-x}}=\frac{2}{\sqrt{1+0}+\sqrt{1-0}}=\frac{2}{1+1}=\frac{2}{2}=1$$

Therefore, the correct answer is 1.

Answer 2

Use the first-order binomial expansion for square root around \( x = 0 \): \[ \sqrt{1 + x} \approx 1 + \frac{x}{2}, \quad \sqrt{1 - x} \approx 1 - \frac{x}{2} \] Now substitute in the limit: \[ \lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x} = \lim_{x \to 0} \frac{\left(1 + \frac{x}{2}\right) - \left(1 - \frac{x}{2}\right)}{x} = \lim_{x \to 0} \frac{\frac{x}{2} + \frac{x}{2}}{x} = \frac{x}{x} = \boxed{1} \] Alternatively, rationalize the numerator: \[ \frac{\sqrt{1+x} - \sqrt{1-x}}{x} \cdot \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} = \frac{(1+x) - (1 - x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})} \] \[ = \frac{2}{\sqrt{1+x} + \sqrt{1-x}} \xrightarrow{x \to 0} \frac{2}{1 + 1} = \boxed{1} \]