Question

\\\lim_{x \to \infty} \frac{\sqrt{2x^2 + 1}}{3x - 5}\\

Explanation:

Identify the limit expression

We need to evaluate the limit:

$$ L = \lim_{x \to \infty} \frac{\sqrt{2x^2 + 1}}{3x - 5} $$

Divide by the highest power

Using the Infinite Limits concept, we divide the numerator and denominator by \(x\). Since \(x \to \infty\), we have \(x > 0\), so \(x = \sqrt{x^2}\).

$$ L = \lim_{x \to \infty} \frac{\frac{\sqrt{2x^2 + 1}}{x}}{\frac{3x - 5}{x}} $$

Simplify the algebraic terms

Rewrite the expression inside the square root and the denominator:

$$ L = \lim_{x \to \infty} \frac{\sqrt{\frac{2x^2 + 1}{x^2}}}{3 - \frac{5}{x}} = \lim_{x \to \infty} \frac{\sqrt{2 + \frac{1}{x^2}}}{3 - \frac{5}{x}} $$

Evaluate the limit of each term

As \(x \to \infty\), the terms \(\frac{1}{x^2} \to 0\) and \(\frac{5}{x} \to 0\):

$$ L = \frac{\sqrt{2 + 0}}{3 - 0} = \frac{\sqrt{2}}{3} $$

Answer:

$$\frac{\sqrt{2}}{3}$$