Computing Limits — Question 7

Question 7 - Solution

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

Step 1: Identify the Indeterminate Form

Substituting $x = 0$ gives $$\frac{\sqrt{1} - \sqrt{1}}{0} = \frac{0}{0},$$ which is an indeterminate form. Algebraic simplification is required.

Step 2: Multiply by the Conjugate

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 $$\frac{(1 + 3x) - (1 + x)}{x\bigl(\sqrt{1 + 3x} + \sqrt{1 + x}\bigr)} = \frac{2x}{x\bigl(\sqrt{1 + 3x} + \sqrt{1 + x}\bigr)}.$$

Cancel the common factor of $x$: $$\frac{2}{\sqrt{1 + 3x} + \sqrt{1 + x}}.$$

Step 3: Evaluate the Limit

Now substitute $x = 0$: $$\lim_{x \to 0} \frac{2}{\sqrt{1 + 3x} + \sqrt{1 + x}} = \frac{2}{1 + 1} = 1.$$

$$\boxed{1}$$