12 December 2021

Statement

Find the value of the following limit.

$$\lim\limits_{x \to \infty} \sqrt{9x^2 + x} - 3x$$

Solution

$$\lim\limits_{x \to \infty} \sqrt{9x^2 + x} - 3x$$

Let's try to multiply the expression by the conjugate.

$$= \lim\limits_{x \to \infty} (\sqrt{9x^2 + x} - 3x) \times \frac{\sqrt{9x^2 + x} + 3x}{\sqrt{9x^2 + x} + 3x}$$

$$= \lim\limits_{x \to \infty} \frac{(9x^2 + x) - 9x^2}{\sqrt{9x^2 + x} + 3x}$$

$$= \lim\limits_{x \to \infty} \frac{x}{\sqrt{9x^2 + x} + 3x}$$

Since we are analyzing the function when $$x \to \infty$$ we can remove the $$x^2$$ from the square root $$\sqrt{9x^2 + x}$$ and keep the equality.
When $$x > 0$$ we have $$\sqrt{9x^2 + x} = x \sqrt{9 + \frac{1}{x}}$$.

$$= \lim\limits_{x \to \infty} \frac{x}{x \sqrt{9 + \frac{1}{x}} + 3x}$$

$$= \lim\limits_{x \to \infty} \frac{x}{x (\sqrt{9 + \frac{1}{x}} + 3)}$$

$$= \lim\limits_{x \to \infty} \frac{1}{\sqrt{9 + \frac{1}{x}} + 3}$$

$$= \frac{\lim\limits_{x \to \infty} 1}{\lim\limits_{x \to \infty} \sqrt{9 + \frac{1}{x}} + 3}$$

$$= \frac{1}{\sqrt{9 + 0} + 3}$$

$$= \frac{1}{6}$$