The Limit — Question 3

Question 3 - Solution

We are asked to evaluate: $$\lim_{x \to 0} \frac{\sqrt{1 + 3x} - \sqrt{1 - x}}{x}$$

Direct substitution gives an indeterminate form $\frac{0}{0}$, so we simplify.

Multiply the numerator and denominator by the conjugate of the numerator: $$\frac{\sqrt{1 + 3x} - \sqrt{1 - x}}{x} \cdot \frac{\sqrt{1 + 3x} + \sqrt{1 - x}}{\sqrt{1 + 3x} + \sqrt{1 - x}}$$

This yields: $$\frac{(1 + 3x) - (1 - x)}{x\left(\sqrt{1 + 3x} + \sqrt{1 - x}\right)} = \frac{4x}{x\left(\sqrt{1 + 3x} + \sqrt{1 - x}\right)}$$

Cancel $x$: $$\frac{4}{\sqrt{1 + 3x} + \sqrt{1 - x}}$$

Now take the limit as $x \to 0$: $$\frac{4}{\sqrt{1} + \sqrt{1}} = \frac{4}{2} = \boxed{2}$$