Question

consider the following function. f(x) = \frac{5x}{x - 7} describe the local behavior of the function. f(x) is undefined at x = 7. 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 → ∞, f(x) → and as x → -∞, f(x) →. question help: message instructor

Explanation:

Step1: Analyze left - hand limit

We want to find $\lim_{x
ightarrow7^{-}}\frac{5x}{x - 7}$. Let $x=7 - h$, where $h>0$ and $h
ightarrow0$. Then $f(7 - h)=\frac{5(7 - h)}{(7 - h)-7}=\frac{35-5h}{-h}$. As $h
ightarrow0$, $\frac{35-5h}{-h}
ightarrow-\infty$.

Step2: Analyze right - hand limit

We want to find $\lim_{x
ightarrow7^{+}}\frac{5x}{x - 7}$. Let $x = 7+h$, where $h>0$ and $h
ightarrow0$. Then $f(7 + h)=\frac{5(7 + h)}{(7 + h)-7}=\frac{35 + 5h}{h}$. As $h
ightarrow0$, $\frac{35+5h}{h}
ightarrow\infty$.

Step3: Analyze end - behavior

For $\lim_{x
ightarrow\pm\infty}\frac{5x}{x - 7}$, divide both numerator and denominator by $x$: $\lim_{x
ightarrow\pm\infty}\frac{5x/x}{(x - 7)/x}=\lim_{x
ightarrow\pm\infty}\frac{5}{1-\frac{7}{x}}$. As $x
ightarrow\pm\infty$, $\frac{7}{x}
ightarrow0$, so $\lim_{x
ightarrow\pm\infty}\frac{5}{1-\frac{7}{x}} = 5$.

Answer:

As $x$ approaches $7$ from the left side, $f(x)$ approaches $-\infty$.
As $x$ approaches $7$ from the right side, $f(x)$ approaches $\infty$.
As $x
ightarrow\infty$, $f(x)
ightarrow5$.
As $x
ightarrow-\infty$, $f(x)
ightarrow5$.