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