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 \( \lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x} \), rationalize the expression. Multiply the numerator and denominator by the conjugate, \( \sqrt{1 + x} + \sqrt{1 - x} \):

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

Using the difference of squares in the numerator, we get:

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

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

Cancel \( x \) (for \( xeq0 \)):

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

Now, evaluate the limit as \( x \to 0 \):

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

The limit is 1.