Questions: Evaluate the limit as x approaches infinity of (2-(4x^2+x))/(2x^2+1) =

Solution Steps

To evaluate the limit as \( x \) approaches infinity for the given function, we need to analyze the behavior of the numerator and the denominator. We can divide both the numerator and the denominator by \( x^2 \), the highest power of \( x \) in the denominator, and then take the limit as \( x \) approaches infinity.

Given the limit: \[ \lim _{x \rightarrow \infty} \frac{2 - (4x^2 + x)}{2x^2 + 1} \]

First, simplify the numerator: \[ 2 - (4x^2 + x) = -4x^2 - x + 2 \]

So the expression becomes: \[ \frac{-4x^2 - x + 2}{2x^2 + 1} \]

To evaluate the limit as \( x \) approaches infinity, divide both the numerator and the denominator by \( x^2 \), the highest power of \( x \) in the denominator: \[ \frac{\frac{-4x^2 - x + 2}{x^2}}{\frac{2x^2 + 1}{x^2}} = \frac{-4 - \frac{x}{x^2} + \frac{2}{x^2}}{2 + \frac{1}{x^2}} \]

Simplify the fractions: \[ \frac{-4 - \frac{1}{x} + \frac{2}{x^2}}{2 + \frac{1}{x^2}} \]

As \( x \) approaches infinity, the terms \(\frac{1}{x}\) and \(\frac{2}{x^2}\) approach 0: \[ \lim_{x \to \infty} \frac{-4 - \frac{1}{x} + \frac{2}{x^2}}{2 + \frac{1}{x^2}} = \frac{-4 - 0 + 0}{2 + 0} = \frac{-4}{2} = -2 \]

Final Answer