Question

let ( f(x) = \begin{cases} x^2 + 4, & x < -4 \\sqrt{x + 4}, & x ge -4 end{cases} ). compute the following limits or state that they do not exist.

a. ( lim_{x \to -4^-} f(x) )

b. ( lim_{x \to -4^+} f(x) )

c. ( lim_{x \to -4} f(x) )

a. compute the limit of ( lim_{x \to -4^-} 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. ( lim_{x \to -4^-} f(x) = 20 ) (simplify your answer.)
b. the limit does not exist

b. compute the limit of ( lim_{x \to -4^+} 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. ( lim_{x \to -4^+} f(x) = square ) (simplify your answer.)
b. the limit does not exist (selected)

Explanation:

Part b

Step 1: Identify the right - hand limit function

For the right - hand limit as \(x
ightarrow - 4^{+}\), we use the part of the piece - wise function where \(x\geq - 4\). The function is \(f(x)=\sqrt{x + 4}\) for \(x\geq - 4\).

Step 2: Evaluate the limit

To find \(\lim_{x
ightarrow - 4^{+}}f(x)\), we substitute \(x=-4\) into the function \(y = \sqrt{x + 4}\) (since the function \(\sqrt{x + 4}\) is continuous at \(x=-4\) for \(x\geq - 4\)).

We have \(\lim_{x
ightarrow - 4^{+}}\sqrt{x + 4}=\sqrt{-4 + 4}\)

Simplify the expression inside the square root: \(-4 + 4=0\), so \(\sqrt{0}=0\)

Answer:

A. \(\lim_{x
ightarrow - 4^{+}}f(x)=\boldsymbol{0}\) (Simplify your answer)