Question

Find the value of $\displaystyle \lim_{x\to 0} \frac{\sqrt{1+2x} - \sqrt{1-2x}}{\sin x}$

Answer 1

2

Answer 2

Step 1. Expand the square‐roots using the first two terms of the Taylor series at $x=0$:

$$\sqrt{1+2x} \approx 1 + x - \frac{x^2}{2}, \quad \sqrt{1-2x} \approx 1 - x - \frac{x^2}{2}.$$

Step 2. Compute the difference:

$$\sqrt{1+2x} - \sqrt{1-2x} \approx (1 + x - \tfrac{x^2}{2}) - (1 - x - \tfrac{x^2}{2}) = 2x.$$

Step 3. Use $\sin x \approx x$ as $x \to 0$. Hence

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

$\boxed{2}$ is the limit.