Question

find the limits in a) through c) below for the function $f(x)=\frac{x^{2}-8x + 7}{x + 7}$. use $-infty$ and $infty$ when appropriate.
a) select the correct choice below and fill in any answer boxes in your choice
a. $lim_{x
ightarrow - 7^{-}}f(x)=square$ (simplify your answer)
b. the limit does not exist and is neither $-infty$ nor $infty$.
b) select the correct choice below and fill in any answer boxes in your choice
a. $lim_{x
ightarrow - 7^{+}}f(x)=square$ (simplify your answer)
b. the limit does not exist and is neither $-infty$ nor $infty$.
c) select the correct choice below and fill in any answer boxes in your choice
a. $lim_{x
ightarrow - 7}f(x)=square$ (simplify your answer)
b. the limit does not exist and is neither $-infty$ nor $infty$.

Explanation:

Step1: Factor the numerator

We factor \(x^{2}-8x + 7=(x - 1)(x - 7)\). So \(f(x)=\frac{(x - 1)(x - 7)}{x + 7}\).

Step2: Analyze the left - hand limit as \(x\to - 7^{-}\)

As \(x\to - 7^{-}\), the numerator \((x - 1)(x - 7)\to(-7 - 1)(-7 - 7)=(-8)\times(-14)=112\) and the denominator \(x + 7\to0^{-}\). So \(\lim_{x\to - 7^{-}}f(x)=-\infty\).

Step3: Analyze the right - hand limit as \(x\to - 7^{+}\)

As \(x\to - 7^{+}\), the numerator \((x - 1)(x - 7)\to(-7 - 1)(-7 - 7)=112\) and the denominator \(x + 7\to0^{+}\). So \(\lim_{x\to - 7^{+}}f(x)=\infty\).

Step4: Analyze the two - sided limit as \(x\to - 7\)

Since \(\lim_{x\to - 7^{-}}f(x)=-\infty\) and \(\lim_{x\to - 7^{+}}f(x)=\infty\), the two - sided limit \(\lim_{x\to - 7}f(x)\) does not exist and is neither \(-\infty\) nor \(\infty\).

Answer:

a) A. \(\lim_{x\to - 7^{-}}f(x)=-\infty\)
b) A. \(\lim_{x\to - 7^{+}}f(x)=\infty\)
c) B. The limit does not exist and is neither \(-\infty\) nor \(\infty\)