question 19 · 1 point\nfor the function (f(x)) given below, evaluate (lim_{x\rightarrowinfty}f(x)) and…

Explanation:

Step1: Analyze $\lim_{x\rightarrow\infty}f(x)$

For $f(x)=-2x + \sqrt{4x^{2}-2x}$, multiply and divide by the conjugate $-2x-\sqrt{4x^{2}-2x}$. [ \begin{align*} f(x)&=\frac{(-2x + \sqrt{4x^{2}-2x})(-2x-\sqrt{4x^{2}-2x})}{-2x-\sqrt{4x^{2}-2x}}\ &=\frac{4x^{2}-(4x^{2}-2x)}{-2x-\sqrt{4x^{2}-2x}}\ &=\frac{2x}{-2x-\sqrt{4x^{2}-2x}} \end{align*} ] As $x\rightarrow\infty$, divide numerator and denominator by $x$: [ \begin{align*} \lim_{x\rightarrow\infty}\frac{2x}{-2x-\sqrt{4x^{2}-2x}}&=\lim_{x\rightarrow\infty}\frac{2}{-2-\sqrt{4-\frac{2}{x}}}\ &=\frac{2}{-2 - 2}\ &=-\frac{1}{2} \end{align*} ]

Step2: Analyze $\lim_{x\rightarrow-\infty}f(x)$

For $x\rightarrow-\infty$, when we consider $\sqrt{4x^{2}-2x}$, we have $\sqrt{4x^{2}-2x}=|x|\sqrt{4-\frac{2}{x}}=-x\sqrt{4 - \frac{2}{x}}$ (since $x<0$). [ \begin{align*} f(x)&=-2x+\sqrt{4x^{2}-2x}\ &=-2x - x\sqrt{4-\frac{2}{x}}\ &=x(-2-\sqrt{4-\frac{2}{x}}) \end{align*} ] Then $\lim_{x\rightarrow-\infty}f(x)=\lim_{x\rightarrow-\infty}x(-2-\sqrt{4-\frac{2}{x}})=\infty$

Answer:

$\lim_{x\rightarrow\infty}f(x)=-\frac{1}{2}$, $\lim_{x\rightarrow-\infty}f(x)=\infty$