Question

determine the following limits.
a. $limlimits_{x \to -7^-} \frac{(x - 1)(x - 7)}{x + 7}$
b. $limlimits_{x \to -7^+} \frac{(x - 1)(x - 7)}{x + 7}$
c. $limlimits_{x \to -7} \frac{(x - 1)(x - 7)}{x + 7}$
a. select the correct choice and, if necessary, fill in the answer box to complete your choice.
a. $limlimits_{x \to -7^-} \frac{(x - 1)(x - 7)}{x + 7} = square$ (simplify your answer.)
b. the limit does not exist and is neither $infty$ nor $-infty$.
b. select the correct choice and, if necessary, fill in the answer box to complete your choice.
a. $limlimits_{x \to -7^+} \frac{(x - 1)(x - 7)}{x + 7} = square$ (simplify your answer.)
b. the limit does not exist and is neither $infty$ nor $-infty$.
c. select the correct choice and, if necessary, fill in the answer box to complete your choice.
a. $limlimits_{x \to -7} \frac{(x - 1)(x - 7)}{x + 7} = square$ (simplify your answer.)
b. the limit does not exist and is neither $infty$ nor $-infty$.

Explanation:

Part a: $\boldsymbol{\lim_{x \to -7^-} \frac{(x - 1)(x - 7)}{x + 7}}$

Step 1: Analyze the denominator's sign

As $x \to -7^-$, $x$ is slightly less than $-7$, so $x + 7 < 0$ (negative).

Step 2: Analyze the numerator's sign

For $x \to -7^-$, $x - 1 < 0$ (since $x < -7 < 1$) and $x - 7 < 0$ (since $x < -7 < 7$). The product of two negative numbers is positive, so $(x - 1)(x - 7) > 0$.

Step 3: Determine the overall sign and behavior

The fraction is $\frac{\text{positive}}{\text{negative}} = \text{negative}$. Now, we can also consider the limit by factoring or substituting values close to $-7$ from the left. Let's take $x = -7 - h$ where $h \to 0^+$. Then:
\[

$$\begin{align*} \frac{(-7 - h - 1)(-7 - h - 7)}{(-7 - h) + 7} &= \frac{(-8 - h)(-14 - h)}{-h}\\ &= \frac{(8 + h)(14 + h)}{-h} \end{align*}$$

\]
As $h \to 0^+$, the numerator approaches $(8)(14) = 112$ (positive) and the denominator approaches $0$ from the negative side. So the limit is $-\infty$.

Step 1: Analyze the denominator's sign

As $x \to -7^+$, $x$ is slightly greater than $-7$, so $x + 7 > 0$ (positive).

Step 2: Analyze the numerator's sign

For $x \to -7^+$, $x - 1 < 0$ (since $x < -7 < 1$) and $x - 7 < 0$ (since $x < -7 < 7$). The product of two negative numbers is positive, so $(x - 1)(x - 7) > 0$.

Step 3: Determine the overall sign and behavior

The fraction is $\frac{\text{positive}}{\text{positive}} = \text{positive}$. Using the substitution $x = -7 + h$ where $h \to 0^+$:
\[

$$\begin{align*} \frac{(-7 + h - 1)(-7 + h - 7)}{(-7 + h) + 7} &= \frac{(-8 + h)(-14 + h)}{h}\\ &= \frac{(8 - h)(14 - h)}{h} \end{align*}$$

\]
As $h \to 0^+$, the numerator approaches $(8)(14) = 112$ (positive) and the denominator approaches $0$ from the positive side. So the limit is $+\infty$.

Step 1: Compare left and right limits

From part (a), $\lim_{x \to -7^-} \frac{(x - 1)(x - 7)}{x + 7} = -\infty$, and from part (b), $\lim_{x \to -7^+} \frac{(x - 1)(x - 7)}{x + 7} = +\infty$.

Step 2: Determine the existence of the limit

For a limit to exist at a point, the left - hand limit and the right - hand limit must be equal (and finite, or both infinite with the same sign in the case of infinite limits). Here, the left - hand limit is $-\infty$ and the right - hand limit is $+\infty$, so they are not equal.

Answer:

$-\infty$

Part b: $\boldsymbol{\lim_{x \to -7^+} \frac{(x - 1)(x - 7)}{x + 7}}$