Question

consider the following function.

$f(x)=\frac{5x}{x - 4}$

describe the local behavior of the function.

$f(x)$ is undefined at $x=$

• as $x$ approaches this value from the left side,

$f(x)$ approaches

• as $x$ approaches this value from the right side,

$f(x)$ approaches

describe the end behavior of the function.

as $x\to\infty$, $f(x)\to$

and as $x\to-\infty$, $f(x)\to$

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Explanation:

Step1: Analyze the function \( f(x)=\frac{5x}{x - 4}\) for local behavior at \(x = 4\)

• Left - hand limit (\(x\to4^{-}\)):

Let \(x=4 - h\) where \(h\to0^{+}\). Then \(f(x)=\frac{5(4 - h)}{(4 - h)-4}=\frac{20-5h}{-h}\). As \(h\to0^{+}\), \(f(x)\to-\infty\).

• Right - hand limit (\(x\to4^{+}\)):

Let \(x = 4+h\) where \(h\to0^{+}\). Then \(f(x)=\frac{5(4 + h)}{(4 + h)-4}=\frac{20 + 5h}{h}\). As \(h\to0^{+}\), \(f(x)\to+\infty\).

Step2: Analyze the end - behavior as \(x\to\pm\infty\)

• Divide numerator and denominator by \(x\):

\(f(x)=\frac{5x}{x - 4}=\frac{5}{1-\frac{4}{x}}\).

• As \(x\to\infty\):

\(\lim_{x\to\infty}\frac{5}{1-\frac{4}{x}} = 5\) (since \(\lim_{x\to\infty}\frac{4}{x}=0\)).

• As \(x\to-\infty\):

\(\lim_{x\to-\infty}\frac{5}{1-\frac{4}{x}} = 5\) (since \(\lim_{x\to-\infty}\frac{4}{x}=0\)).

Answer:

• As \(x\) approaches \(4\) from the left side, \(f(x)\to-\infty\).
• As \(x\) approaches \(4\) from the right side, \(f(x)\to+\infty\).
• As \(x\to\infty\), \(f(x)\to5\).
• As \(x\to-\infty\), \(f(x)\to5\).