Question

let \\(f(x) = \

$$\begin{cases} x^2 + 19, & x < -19 \\\\ \\sqrt{x + 19}, & x \\ge -19 \\end{cases}$$

\\). compute the following limits or state that they do not exist.

a. \\(\lim_{x \to -19^-} f(x)\\)
b. \\(\lim_{x \to -19^+} f(x)\\)
c. \\(\lim_{x \to -19} f(x)\\)

c. compute the limit of \\(\lim_{x \to -19} f(x)\\) or state that it does not exist. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. yes, \\(\lim_{x \to -19} f(x)\\) exists and equals (simplify your answer.)
b. no, \\(\lim_{x \to -19} f(x)\\) does not exist because \\(\lim_{x \to -19^+} f(x) \
e \lim_{x \to -19^-} f(x)\\)
c. no, \\(\lim_{x \to -19} f(x)\\) does not exist because \\(f(-19)\\) is undefined.

Explanation:

Evaluate the left-hand limit

$$ \lim_{x \to -19^{-}} f(x) = \lim_{x \to -19^{-}} (x^2 + 19) = (-19)^2 + 19 = 361 + 19 = 380 $$

Evaluate the right-hand limit

$$ \lim_{x \to -19^{+}} f(x) = \lim_{x \to -19^{+}} \sqrt{x + 19} = \sqrt{-19 + 19} = 0 $$

Determine the existence of the two-sided limit

$$ \lim_{x \to -19^{-}} f(x) = 380 eq 0 = \lim_{x \to -19^{+}} f(x) \implies \lim_{x \to -19} f(x) \text{ does not exist} $$

Answer:

• A. Yes, \(\lim_{x \to -19} f(x)\) exists and equals ____.
• **B. No, \(\lim_{x \to -19} f(x)\) does not exist because \(\lim_{x \to -19^{+}} f(x)

eq \lim_{x \to -19^{-}} f(x)\). (Correct answer)**

• C. No, \(\lim_{x \to -19} f(x)\) does not exist because \(f(-19)\) is undefined.