- #1

askor

- 169

- 9

$$\lim_{x \to \infty} \frac{4}{\sqrt{x^2 + x} - \sqrt{x^2 - 3x}}$$

This is my attempt to solve the problem:

$$\lim_{x \to \infty} \frac{4}{\sqrt{x^2 + x} - \sqrt{x^2 - 3x}}$$

$$= \lim_{x \to \infty} \frac{4}{\sqrt{x^2 + x} - \sqrt{x^2 - 3x}} × \frac{\sqrt{x^2 + x} + \sqrt{x^2 - 3x}}{\sqrt{x^2 + x} + \sqrt{x^2 - 3x}}$$

$$= \lim_{x \to \infty} \frac{4(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{(x^2 + x) - (x^2 - 3x)}$$

$$= \lim_{x \to \infty} \frac{4(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{x^2 + x - x^2 + 3x}$$

$$= \lim_{x \to \infty} \frac{4(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{4x}$$

$$= \lim_{x \to \infty} \frac{(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{x}$$

$$= \lim_{x \to \infty} \frac{\frac{1}{\sqrt{x^2}}(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{\frac{1}{\sqrt{x^2}}(x)}$$

$$= \lim_{x \to \infty} \frac{\sqrt{\frac{x^2 + x}{x^2}} + \sqrt{\frac{x^2 - 3x}{x^2}}}{\frac{x}{x}}$$

$$= \lim_{x \to \infty} \frac{\sqrt{1 + \frac{1}{x}} + \sqrt{1 - \frac{3}{x}}}{1}$$

$$= \sqrt{1 + \frac{1}{\infty}} + \sqrt{1 - \frac{3}{\infty}}$$

$$= \sqrt{1 + 0} + \sqrt{1 - 0}$$

$$= \sqrt{1} + \sqrt{1}$$

$$= 1 + 1$$

$$= 2$$

Is this correct?