Question

evaluate the limit. leave results in exact form.

\\\lim_{x \to \infty} \frac{\sqrt{4 + 8x^2}}{6 + 4x}\\

Explanation:

Divide numerator and denominator by \(x\)

$$ \lim_{x \to \infty} \frac{\sqrt{4 + 8x^2}}{6 + 4x} = \lim_{x \to \infty} \frac{\frac{\sqrt{4 + 8x^2}}{x}}{\frac{6 + 4x}{x}} $$

Simplify the algebraic expressions

Since \(x > 0\) as \(x \to \infty\), we have \(x = \sqrt{x^2}\):

$$ \lim_{x \to \infty} \frac{\sqrt{\frac{4}{x^2} + 8}}{\frac{6}{x} + 4} $$

Evaluate the limit of each term

As \(x \to \infty\), \(\frac{4}{x^2} \to 0\) and \(\frac{6}{x} \to 0\):

$$ \frac{\sqrt{0 + 8}}{0 + 4} = \frac{\sqrt{8}}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} $$

Answer:

Evaluate the limit. Leave results in exact form.

$$\lim_{x\to\infty} \frac{\sqrt{4 + 8x^2}}{6 + 4x}$$

<blank>\(\frac{\sqrt{2}}{2}\)</blank>