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 \(\sqrt{9x^2 + x} = x \sqrt{9 + \frac{1}{x}}\).

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

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

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

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

\(= \frac{1}{\sqrt{9 + 0} + 3}\)

\(= \frac{1}{6}\)