Question

directions: for each of the following, write the left and right limit statements for f(x) as x approaches 3.

1. $f(x)=\frac{(x + 2)(x - 6)}{x - 3}$

left: $lim_{x
ightarrow3^{-}}f(x)=$
right: $lim_{x
ightarrow3^{+}}$

1. $f(x)=\frac{(x - 3)(x + 3)}{x(x - 3)}$

left:
right:

1. $f(x)=\frac{-2}{(x - 3)^2}$

left:
right:

Explanation:

Step1: Recall left - hand limit definition

The left - hand limit as $x\to a^-$ is $\lim_{x\to a^-}f(x)$. We substitute values of $x$ that are less than $a$ and approach $a$.

Step2: Recall right - hand limit definition

The right - hand limit as $x\to a^+$ is $\lim_{x\to a^+}f(x)$. We substitute values of $x$ that are greater than $a$ and approach $a$.

For $f(x)=\frac{(x + 2)(x - 6)}{x-3}$:

Left - hand limit

As $x\to3^-$, we consider values of $x$ slightly less than 3. Let $x = 3 - h$, where $h>0$ and $h\to0$. Then $f(3 - h)=\frac{(3 - h+2)(3 - h - 6)}{(3 - h)-3}=\frac{(5 - h)(-3 - h)}{-h}=\frac{-15-5h + 3h+h^{2}}{-h}=\frac{h^{2}-2h - 15}{-h}$. As $h\to0$, $\lim_{x\to3^-}f(x)=+\infty$.

Right - hand limit

As $x\to3^+$, we consider values of $x$ slightly greater than 3. Let $x = 3+h$, where $h>0$ and $h\to0$. Then $f(3 + h)=\frac{(3 + h+2)(3 + h - 6)}{(3 + h)-3}=\frac{(5 + h)(-3 + h)}{h}=\frac{-15 + 5h-3h+h^{2}}{h}=\frac{h^{2}+2h - 15}{h}$. As $h\to0$, $\lim_{x\to3^+}f(x)=-\infty$.

For $f(x)=\frac{(x - 3)(x + 3)}{x(x - 3)}$ (simplify to $f(x)=\frac{x + 3}{x}$ for $x

eq3$):

Left - hand limit

As $x\to3^-$, $f(x)=\frac{x + 3}{x}$. Let $x = 3 - h$, where $h>0$ and $h\to0$. Then $f(3 - h)=\frac{3 - h+3}{3 - h}=\frac{6 - h}{3 - h}$. As $h\to0$, $\lim_{x\to3^-}f(x)=\frac{6-0}{3 - 0}=2$.

Right - hand limit

As $x\to3^+$, let $x = 3+h$, where $h>0$ and $h\to0$. Then $f(3 + h)=\frac{3 + h+3}{3 + h}=\frac{6 + h}{3 + h}$. As $h\to0$, $\lim_{x\to3^+}f(x)=2$.

For $f(x)=\frac{-2}{(x - 3)^{2}}$:

Left - hand limit

As $x\to3^-$, let $x = 3 - h$, where $h>0$ and $h\to0$. Then $f(3 - h)=\frac{-2}{(3 - h-3)^{2}}=\frac{-2}{h^{2}}$. As $h\to0$, $\lim_{x\to3^-}f(x)=-\infty$.

Right - hand limit

As $x\to3^+$, let $x = 3+h$, where $h>0$ and $h\to0$. Then $f(3 + h)=\frac{-2}{(3 + h-3)^{2}}=\frac{-2}{h^{2}}$. As $h\to0$, $\lim_{x\to3^+}f(x)=-\infty$.

Answer:

• For $f(x)=\frac{(x + 2)(x - 6)}{x-3}$: Left: $\lim_{x\to3^-}f(x)=+\infty$, Right: $\lim_{x\to3^+}f(x)=-\infty$
• For $f(x)=\frac{(x - 3)(x + 3)}{x(x - 3)}$: Left: $\lim_{x\to3^-}f(x)=2$, Right: $\lim_{x\to3^+}f(x)=2$
• For $f(x)=\frac{-2}{(x - 3)^{2}}$: Left: $\lim_{x\to3^-}f(x)=-\infty$, Right: $\lim_{x\to3^+}f(x)=-\infty$