Questions: Evaluate the limit as x approaches infinity of (sqrt(9x^2+2x)-3x). 1 / 3 0 2 / 3 infinity

Solution Steps

To evaluate the limit \(\lim _{x \rightarrow \infty}\left(\sqrt{9 x^{2}+2 x}-3 x\right)\), we can factor out \(x\) from the square root to simplify the expression. This involves rewriting the expression inside the square root in a form that allows us to factor out \(x^2\), and then simplifying the resulting expression as \(x\) approaches infinity.

We start with the limit expression: \[ \lim_{x \rightarrow \infty} \left(\sqrt{9x^2 + 2x} - 3x\right) \] To simplify, we factor out \(x^2\) from the square root: \[ \sqrt{9x^2 + 2x} = \sqrt{x^2(9 + \frac{2}{x})} = x\sqrt{9 + \frac{2}{x}} \] Thus, the expression becomes: \[ \lim_{x \rightarrow \infty} \left(x\sqrt{9 + \frac{2}{x}} - 3x\right) \]

We can factor out \(x\) from the limit: \[ \lim_{x \rightarrow \infty} x\left(\sqrt{9 + \frac{2}{x}} - 3\right) \] As \(x\) approaches infinity, \(\frac{2}{x}\) approaches \(0\), so we have: \[ \sqrt{9 + \frac{2}{x}} \rightarrow \sqrt{9} = 3 \] Thus, the expression simplifies to: \[ \lim_{x \rightarrow \infty} x\left(3 - 3\right) = \lim_{x \rightarrow \infty} x \cdot 0 = 0 \]

To find the limit more accurately, we can use the expansion: \[ \sqrt{9 + \frac{2}{x}} \approx 3 + \frac{1}{3}\cdot\frac{2}{x} \quad \text{(using Taylor expansion)} \] Substituting this back, we get: \[ \lim_{x \rightarrow \infty} x\left(\left(3 + \frac{2}{3x}\right) - 3\right) = \lim_{x \rightarrow \infty} x\cdot\frac{2}{3x} = \frac{2}{3} \]

Final Answer