Question

let f(x) = { (x^2 - 144)/(x + 12) if x < 0, (x^2 - 144)/(x - 12) if x > 0 }. find a) lim(x→ - 12) f(x), b) lim(x→0) f(x), and c) lim(x→12) f(x). a) select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. lim(x→ - 12) f(x) = (simplify your answer.) b. the limit does not exist.

Explanation:

Step1: Simplify the function for \(x < 0\)

When \(x<0\), \(f(x)=\frac{x^{2}-144}{x + 12}\). Using the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), where \(a=x\) and \(b = 12\), we have \(f(x)=\frac{(x + 12)(x - 12)}{x+12}=x - 12\) for \(x
eq - 12\).

Step2: Find \(\lim_{x

ightarrow - 12}f(x)\)
Since \(x
ightarrow - 12\) and \(x<0\), we use the simplified form \(f(x)=x - 12\). Substitute \(x=-12\) into \(x - 12\), we get \(\lim_{x
ightarrow - 12}f(x)=-12-12=-24\).

Step3: Analyze \(\lim_{x

ightarrow0}f(x)\)
We need to find the left - hand limit \(\lim_{x
ightarrow0^{-}}f(x)\) and the right - hand limit \(\lim_{x
ightarrow0^{+}}f(x)\).
For \(x
ightarrow0^{-}\), \(f(x)=\frac{x^{2}-144}{x + 12}\), and \(\lim_{x
ightarrow0^{-}}f(x)=\frac{0^{2}-144}{0 + 12}=-12\).
For \(x
ightarrow0^{+}\), \(f(x)=\frac{x^{2}-144}{x - 12}\), and \(\lim_{x
ightarrow0^{+}}f(x)=\frac{0^{2}-144}{0 - 12}=12\). Since \(\lim_{x
ightarrow0^{-}}f(x)
eq\lim_{x
ightarrow0^{+}}f(x)\), \(\lim_{x
ightarrow0}f(x)\) does not exist.

Step4: Analyze \(\lim_{x

ightarrow12}f(x)\)
Since \(x
ightarrow12\) and \(x>0\), \(f(x)=\frac{x^{2}-144}{x - 12}=\frac{(x + 12)(x - 12)}{x - 12}=x + 12\) for \(x
eq12\). Substitute \(x = 12\) into \(x + 12\), we get \(\lim_{x
ightarrow12}f(x)=12+12=24\).

Answer:

a) A. \(\lim_{x
ightarrow - 12}f(x)=-24\)
b) The limit does not exist.
c) \(\lim_{x
ightarrow12}f(x)=24\)