Question

3 multiple choice 1 point evaluate the limit, if it exists. \\(\lim\limits_{x \to \infty} \frac{6x^2 + 7x - 9}{5x^2 + 2}\\) \\(\infty\\) \\(\frac{6}{5}\\) 0 \\(-\frac{2}{9}\\)

Explanation:

Step1: Divide numerator and denominator by \(x^2\)

For the limit \(\lim_{x \to \infty} \frac{6x^2 + 7x - 9}{5x^2 + 2}\), divide each term in the numerator and the denominator by \(x^2\) (the highest power of \(x\) in the denominator).
We get:

$$ \lim_{x \to \infty} \frac{\frac{6x^2}{x^2} + \frac{7x}{x^2} - \frac{9}{x^2}}{\frac{5x^2}{x^2} + \frac{2}{x^2}} $$

Step2: Simplify each term

Simplify each fraction:

$$ \lim_{x \to \infty} \frac{6 + \frac{7}{x} - \frac{9}{x^2}}{5 + \frac{2}{x^2}} $$

Step3: Evaluate the limit as \(x \to \infty\)

As \(x \to \infty\), terms with \(\frac{1}{x}\) or \(\frac{1}{x^2}\) approach \(0\) (since \(\lim_{x \to \infty} \frac{1}{x} = 0\) and \(\lim_{x \to \infty} \frac{1}{x^2} = 0\)).
Substitute these limits:

$$ \frac{6 + 0 - 0}{5 + 0} = \frac{6}{5} $$

Answer:

\(\frac{6}{5}\) (corresponding to the option with \(\boldsymbol{\frac{6}{5}}\))