Question

analyze the following limits and find the vertical asymptotes of ( f(x) = \frac{x - 12}{x^2 - 144} ).

a. ( limlimits_{x \to 12} f(x) )

b. ( limlimits_{x \to -12^-} f(x) )

c. ( limlimits_{x \to -12^+} f(x) )

( limlimits_{x \to 12} \frac{x - 12}{x^2 - 144} = limlimits_{x \to 12} left( \frac{1}{x + 12}
ight) ) (simplify your answer.)

now, evaluate the limit. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. ( limlimits_{x \to 12} f(x) = \frac{1}{24} ) (simplify your answer.)

b. the limit does not exist and is neither ( infty ) nor ( -infty ).

b. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. ( limlimits_{x \to -12^-} f(x) = ) (simplify your answer.)

b. the limit does not exist and is neither ( infty ) nor ( -infty ).

Explanation:

Part b: $\boldsymbol{\lim_{x \to -12^-} f(x)}$

First, simplify $f(x)$:
The denominator $x^2 - 144$ factors as $(x - 12)(x + 12)$, so:

$$f(x) = \frac{x - 12}{(x - 12)(x + 12)} = \frac{1}{x + 12} \quad (x eq 12)$$

Now, analyze the limit as $x \to -12^-$ (approaching $-12$ from the left):

• Let $x = -12 - h$ where $h \to 0^+$ (small positive number).
• Substitute into $\frac{1}{x + 12}$:

$$\frac{1}{(-12 - h) + 12} = \frac{1}{-h}$$

• As $h \to 0^+$, $-h \to 0^-$, so $\frac{1}{-h} \to -\infty$.

Part c: $\boldsymbol{\lim_{x \to -12^+} f(x)}$

Analyze the limit as $x \to -12^+$ (approaching $-12$ from the right):

• Let $x = -12 + h$ where $h \to 0^+$ (small positive number).
• Substitute into $\frac{1}{x + 12}$:

$$\frac{1}{(-12 + h) + 12} = \frac{1}{h}$$

• As $h \to 0^+$, $\frac{1}{h} \to +\infty$.

Vertical Asymptotes

A vertical asymptote occurs where the function has an infinite limit (or the denominator is zero and the numerator is non - zero).

• For $x = 12$: We saw that $\lim_{x \to 12} f(x)=\frac{1}{24}$ (a finite limit), so $x = 12$ is a removable discontinuity (a hole), not a vertical asymptote.
• For $x=-12$: Since $\lim_{x \to -12^-} f(x)=-\infty$ and $\lim_{x \to -12^+} f(x)=+\infty$, the function has a vertical asymptote at $x=-12$.

Final Answers

• a. $\lim_{x \to 12} f(x)=\boldsymbol{\frac{1}{24}}$
• b. $\lim_{x \to -12^-} f(x)=\boldsymbol{-\infty}$ (the limit does not exist in the real - number sense, it approaches $-\infty$)
• c. $\lim_{x \to -12^+} f(x)=\boldsymbol{+\infty}$ (the limit does not exist in the real - number sense, it approaches $+\infty$)
• Vertical Asymptote: $\boldsymbol{x=-12}$

Answer:

Part b: $\boldsymbol{\lim_{x \to -12^-} f(x)}$

First, simplify $f(x)$:
The denominator $x^2 - 144$ factors as $(x - 12)(x + 12)$, so:

$$f(x) = \frac{x - 12}{(x - 12)(x + 12)} = \frac{1}{x + 12} \quad (x eq 12)$$

Now, analyze the limit as $x \to -12^-$ (approaching $-12$ from the left):

• Let $x = -12 - h$ where $h \to 0^+$ (small positive number).
• Substitute into $\frac{1}{x + 12}$:

$$\frac{1}{(-12 - h) + 12} = \frac{1}{-h}$$

• As $h \to 0^+$, $-h \to 0^-$, so $\frac{1}{-h} \to -\infty$.

Part c: $\boldsymbol{\lim_{x \to -12^+} f(x)}$

Analyze the limit as $x \to -12^+$ (approaching $-12$ from the right):

• Let $x = -12 + h$ where $h \to 0^+$ (small positive number).
• Substitute into $\frac{1}{x + 12}$:

$$\frac{1}{(-12 + h) + 12} = \frac{1}{h}$$

• As $h \to 0^+$, $\frac{1}{h} \to +\infty$.

Vertical Asymptotes

A vertical asymptote occurs where the function has an infinite limit (or the denominator is zero and the numerator is non - zero).

• For $x = 12$: We saw that $\lim_{x \to 12} f(x)=\frac{1}{24}$ (a finite limit), so $x = 12$ is a removable discontinuity (a hole), not a vertical asymptote.
• For $x=-12$: Since $\lim_{x \to -12^-} f(x)=-\infty$ and $\lim_{x \to -12^+} f(x)=+\infty$, the function has a vertical asymptote at $x=-12$.

Final Answers

• a. $\lim_{x \to 12} f(x)=\boldsymbol{\frac{1}{24}}$
• b. $\lim_{x \to -12^-} f(x)=\boldsymbol{-\infty}$ (the limit does not exist in the real - number sense, it approaches $-\infty$)
• c. $\lim_{x \to -12^+} f(x)=\boldsymbol{+\infty}$ (the limit does not exist in the real - number sense, it approaches $+\infty$)
• Vertical Asymptote: $\boldsymbol{x=-12}$