Question

question 19 · 1 point
for the function (f(x)) given below, evaluate (lim_{x
ightarrowinfty}f(x)) and (lim_{x
ightarrow-infty}f(x)).
(f(x)=-2x + sqrt{4x^{2}-2x})
provide your answer below:
(lim_{x
ightarrowinfty}f(x)=)
(lim_{x
ightarrow-infty}f(x)=)

Explanation:

Step1: Analyze $\lim_{x

ightarrow\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 $x
ightarrow\infty$, divide numerator and denominator by $x$:
\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{2x}{-2x-\sqrt{4x^{2}-2x}}&=\lim_{x ightarrow\infty}\frac{2}{-2-\sqrt{4-\frac{2}{x}}}\\ &=\frac{2}{-2 - 2}\\ &=-\frac{1}{2} \end{align*}$$

\]

Step2: Analyze $\lim_{x

ightarrow-\infty}f(x)$
For $x
ightarrow-\infty$, 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 $\lim_{x
ightarrow-\infty}f(x)=\lim_{x
ightarrow-\infty}x(-2-\sqrt{4-\frac{2}{x}})=\infty$

Answer:

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