Question

given the function
$f(x) \frac{(x - 2)(x + 4)}{x(x + 2)}$
what is true about its behavior
near 0

Explanation:

Step1: Simplify the function

$$\begin{align*} f(x)&=\frac{(x - 2)(x + 4)}{x(x + 2)}\\ &=\frac{x^{2}+4x-2x - 8}{x(x + 2)}\\ &=\frac{x^{2}+2x-8}{x(x + 2)} \end{align*}$$

Step2: Analyze the limit as \(x\to0\)

We use the fact that \(\lim_{x
ightarrow0}f(x)=\lim_{x
ightarrow0}\frac{(x - 2)(x + 4)}{x(x + 2)}\). Substitute \(x = 0\) into the non - zero factors: \(\lim_{x
ightarrow0}\frac{(x - 2)(x + 4)}{x(x + 2)}=\lim_{x
ightarrow0}\frac{(-2)(4)}{x(2)}=\lim_{x
ightarrow0}\frac{- 8}{2x}=\lim_{x
ightarrow0}\frac{-4}{x}\)

As \(x\to0^{+}\), \(\lim_{x
ightarrow0^{+}}\frac{-4}{x}=-\infty\)

As \(x\to0^{-}\), \(\lim_{x
ightarrow0^{-}}\frac{-4}{x}=\infty\)

Answer:

As \(x\) approaches \(0\) from the right (\(x\to0^{+}\)), \(f(x)\to-\infty\) and as \(x\) approaches \(0\) from the left (\(x\to0^{-}\)), \(f(x)\to\infty\)